📘 Concept & Theory Theory and Concept Used ›
In this question, we use one of the most important properties of circles:
“The angle subtended by an arc at the centre of a circle is twice the angle subtended by it at any point on the remaining part of the circle.”
Here, arc \(\small AC\) subtends:
- Central angle \(\small \angle AOC\)
- Angle at the circumference \(\small \angle ADC\)
Therefore,
\[\small \angle AOC = 2\angle ADC \]
This theorem is extremely important for solving geometry questions involving:
- Angles in the same segment
- Cyclic quadrilaterals
- Arc-angle relationships
- Olympiad and entrance examination geometry problems
🗺️ Solution Roadmap Step-by-step Plan ›
Find the central angle \(\small \angle AOC\) by adding \(\small \angle AOB\) and \(\small \angle BOC\).
Use the theorem:
Angle at centre \(\small =\) Twice the angle at the circumference.Calculate \(\small \angle ADC\).
📊 Graph / Figure Graph / Figure ›
✏️ Solution Complete Solution ›
- Given
- \[\small \angle BOC = 30^\circ \]
- \[\small \angle AOB = 60^\circ \]
- \(\small D\) is a point on the remaining part of the circle.
- To Find
- \[\small \angle ADC\]
- Step-by-step Solution
- First, find the angle subtended by arc \(\small AC\) at the centre.
- Since point \(\small B\) lies between \(\small A\) and \(\small C\),
- \[\small \angle AOC = \angle AOB + \angle BOC\]
- Substituting the given values: \[\small \begin{aligned}\angle AOC &= 60^\circ + 30^\circ\\\angle AOC &= 90^\circ\end{aligned}\]
- Now, by the theorem:\[\small \text{Angle at the centre} = 2 \times \text{Angle at the circumference}\]
- Therefore,\[\small \angle AOC = 2\angle ADC\]
- Substituting the value of \(\small \angle AOC\):\[\small 90^\circ = 2\angle ADC\]
- Dividing both sides by \(\small 2\),
- \[\small \begin{aligned}\angle ADC &= \frac{90^\circ}{2}\\\angle ADC &= 45^\circ\end{aligned}\]
🎯 Exam Significance Exam Significance ›
This problem is highly important because it strengthens the understanding of the relationship between central angles and angles at the circumference.
Questions based on this theorem are frequently asked in:
- CBSE Board Examinations
- School annual examinations
- NTSE Foundation preparation
- Olympiad geometry problems
- JEE and other entrance exam foundation geometry
Students often lose marks by directly writing the answer without mentioning the theorem used. Writing the theorem explicitly improves presentation and scoring in board examinations.
🔑 Key Takeaways Key Takeaways ›
-
The angle subtended at the centre is twice the angle subtended at the circumference by the same arc.
-
Always identify the common arc before applying circle theorems.
-
Central angles are generally easier to calculate first.
-
Proper theorem statement improves mathematical presentation.
-
Circle geometry theorems form the foundation for higher geometry topics.
📘 Concept & Theory Theory and Concept ›
This question is based on important circle theorems and properties of triangles.
Since the chord is equal to the radius of the circle:
\[\small AB = OA = OB \]
Therefore, \(\small \triangle OAB\) becomes an equilateral triangle.
In an equilateral triangle:
\[\small \text{Each angle} = 60^\circ \]
We also use the theorem:
“The angle subtended by an arc at the centre is twice the angle subtended at any point on the remaining part of the circle.”
Another important fact used is:
Angles in opposite segments formed by the same chord are supplementary.
🗺️ Solution Roadmap Step-by-step Plan ›
Prove that \(\small \triangle OAB\) is an equilateral triangle.
Find the central angle \(\small \angle AOB\).
Find the angle subtended at a point on the major arc using the central angle theorem.
Find the angle subtended at a point on the minor arc using supplementary angles.
📊 Graph / Figure Graph / Figure ›
✏️ Solution Complete Solution ›
- Let the centre of the circle be \(\small O\).
- Let \(\small AB\) be the chord such that:
- AB = \(\small \text{Radius of the circle}\)
- Since \(\small OA\) and \(\small OB\) are also radii of the circle,\[\small OA = OB = AB\]
- Therefore, all three sides of \(\small \triangle OAB\) are equal.
- Hence, \(\small \triangle OAB\) is an equilateral triangle.
- In an equilateral triangle, all angles are equal to \(\small 60^\circ\).
- Therefore,\[\small \angle AOB = 60^\circ\]
- Now, let \(\small D\) be a point on the major arc \(\small AB\).
- The angle subtended by chord \(\small AB\) at point \(\small D\) lies on the remaining part of the circle.
- Using the theorem:\[\small \text{Angle at the centre} = 2 \times \text{Angle at the circumference}\]
- Therefore,\[\small \begin{aligned}\angle ADB &= \frac{1}{2}\angle AOB\\ &= \frac{1}{2}\times 60^\circ\\&= 30^\circ\end{aligned}\]
- Thus, the angle subtended by the chord at a point on the major arc is:\[\small 30^\circ\]
- Now, let \(\small C\) be a point on the minor arc \(\small AB\).
- Angles subtended by the same chord in opposite segments are supplementary.
- Therefore,\[\small \angle ACB + \angle ADB = 180^\circ\]
- Substituting the value of \(\small \angle ADB\):\[\small \begin{aligned} \angle ACB + 30^\circ &= 180^\circ\\ \angle ACB &= 180^\circ - 30^\circ\\ \angle ACB &= 150^\circ\end{aligned}\]
- Thus, the angle subtended by the chord at a point on the minor arc is:\[\small 150^\circ\]
🎯 Exam Significance Exam Significance ›
This problem is highly important because it combines:
- Properties of equilateral triangles
- Circle theorems
- Angles in major and minor segments
Such mixed-concept geometry questions are frequently asked in:
- CBSE Board Examinations
- Foundation mathematics tests
- NTSE and Olympiad examinations
- JEE Foundation geometry preparation
Students should carefully identify whether the point lies on the major arc or minor arc because this changes the angle completely.
🔑 Key Takeaways Key Takeaways ›
-
If a chord equals the radius, the triangle formed with the centre becomes equilateral.
-
Every angle of an equilateral triangle is \(\small 60^\circ\).
-
The angle at the centre is twice the angle at the circumference.
-
Angles in opposite segments are supplementary.
-
Major arc angles are smaller while minor arc angles are larger for the same chord.
📘 Concept & Theory Theory and Concept ›
This problem uses one of the most important circle theorems:
The angle subtended by an arc at the centre is twice the angle subtended by it at any point on the remaining part of the circle.
Since \(\small \angle PQR\) lies on the circumference and subtends arc \(\small PR\), the central angle subtended by the same arc is:
\[\small \angle POR = 2\angle PQR \]
Another important concept used here is:
- Interior angle and reflex angle together make \(\small 360^\circ\).
- Radii of a circle are equal.
- A triangle having two equal sides is an isosceles triangle.
🗺️ Solution Roadmap Step-by-step Plan ›
Find the reflex central angle subtended by chord \(\small PR\).
Convert the reflex angle into the interior angle of triangle \(\small POR\).
Use the fact that \(\small OP = OR\) to identify an isosceles triangle.
Apply the angle sum property of a triangle to find \(\small \angle OPR\).
📊 Graph / Figure Graph / Figure ›
✏️ Solution Complete Solution ›
- Given
- \[\small \angle PQR = 100^\circ\] Points \(\small P\), \(\small Q\), and \(\small R\) lie on a circle with centre \(\small O\).
- We have to find:\[\small \angle OPR\]
- The angle subtended by chord \(\small PR\) at the centre is twice the angle subtended at the remaining part of the circle.
- Therefore,\[\small \begin{aligned}\angle POR &= 2 \times \angle PQR\\&= 2 \times 100^\circ\\&= 200^\circ \end{aligned}\]
- This is the reflex angle subtended at the centre.
- But in \(\small \triangle POR\), we require the interior angle at \(\small O\).
- Therefore,\[\small \begin{aligned} \text{Interior } \angle POR &= 360^\circ - 200^\circ\\ &= 160^\circ \end{aligned}\]
- Since \(\small OP\) and \(\small OR\) are radii of the same circle,\[\small OP = OR\]
- Therefore, \(\small \triangle POR\) is an isosceles triangle.
- Hence,\[\small \angle OPR = \angle ORP\]
- Let:\[\small \angle OPR = \angle ORP = x\]
- Using the angle sum property of a triangle:\[\small \begin{aligned}x + x + 160^\circ &= 180^\circ\\2x + 160^\circ &= 180^\circ\\2x &= 20^\circ \\x&=10^\circ\end{aligned}\]
- Therefore,\[\small \angle OPR = 10^\circ\]
🎯 Exam Significance Exam Significance ›
This problem is very important because it tests multiple geometric concepts together.
Students must carefully understand:
- Difference between reflex angle and interior angle
- Central angle theorem
- Properties of isosceles triangles
- Triangle angle sum property
Such integrated geometry problems are frequently asked in:
- CBSE Board Examinations
- NTSE examinations
- Mathematics Olympiads
- JEE Foundation geometry preparation
A common mistake is directly using \(\small 200^\circ\) inside the triangle instead of converting it to the interior angle \(\small 160^\circ\).
🔑 Key Takeaways Key Takeaways ›
-
Angle at the centre is twice the angle at the circumference.
-
Reflex angle and interior angle together form \(\small 360^\circ\).
-
Radii of a circle are always equal.
-
Equal sides in a triangle imply equal opposite angles.
-
Always check whether the central angle obtained is reflex or interior.
📘 Concept & Theory Theory and Concept ›
This question is based on two important geometry concepts:
- Angle sum property of a triangle
- Angles in the same segment of a circle are equal
In any triangle:
\[\small \text{Sum of all interior angles} = 180^\circ \]
Also, when two angles subtend the same chord of a circle, they are equal if they lie in the same segment.
Here, both \(\small \angle BAC\) and \(\small \angle BDC\) subtend the same chord \(\small BC\).
🗺️ Solution Roadmap Step-by-step Plan ›
Use the angle sum property in \(\small \triangle ABC\) to find \(\small \angle BAC\).
Identify the common chord subtended by the two angles.
Apply the theorem:
Angles in the same segment of a circle are equal.Find \(\small \angle BDC\).
📊 Graph / Figure Graph / Figure ›
✏️ Solution Complete Solution ›
- Given
- \[\small \angle ABC = 69^\circ\] \[\small \angle ACB = 31^\circ\]
- To Find
- \[\small \angle BDC\]
- First, consider \(\small \triangle ABC\).
- By the angle sum property of a triangle:\[\small \angle ABC + \angle ACB + \angle BAC = 180^\circ\]
- Substituting the given values: \[\small \begin{aligned} 69^\circ + 31^\circ + \angle BAC &= 180^\circ\\ 100^\circ + \angle BAC &= 180^\circ\\ \angle BAC &= 80^\circ \end{aligned} \]
- Now observe that:\(\small \angle BAC\) subtends chord \(\small BC\)
- \(\small \angle BDC\) also subtends the same chord \(\small BC\)
- By the theorem: Angles in the same segment of a circle are equal.
- Therefore, \[\small \angle BDC = \angle BAC\]
- \[\small \angle BDC = 80^\circ\]
🎯 Exam Significance Exam Significance ›
This problem is very important because it connects algebraic angle calculation with geometric circle theorems.
Students learn how to:
- Use triangle angle properties efficiently
- Identify common chords in circle geometry
- Apply the same segment theorem correctly
Questions based on the same segment theorem are frequently asked in:
- CBSE Board Examinations
- School term examinations
- NTSE and Olympiad examinations
- JEE Foundation geometry preparation
Students commonly make mistakes by applying the theorem without identifying the common chord first.
🔑 Key Takeaways Key Takeaways ›
-
Sum of angles in a triangle is always \(\small 180^\circ\).
-
Angles subtended by the same chord in the same segment are equal.
-
Always identify the common chord before applying circle theorems.
-
Circle geometry often combines multiple concepts together.
-
Proper theorem statement improves board exam presentation.
📘 Concept & Theory Theory and Concept Used ›
This problem combines properties of intersecting lines and circle theorems.
The following concepts are used:
- Vertically opposite angles are equal.
- Adjacent angles on a straight line form \(\small 180^\circ\).
- Angles subtended by the same chord in the same segment are equal.
- Sum of angles in a triangle is \(\small 180^\circ\).
Here, \(\small \angle ECD\) and \(\small \angle EBA\) subtend the same chord \(\small AD\), therefore they are equal.
🗺️ Solution Roadmap Step-by-step Plan ›
Use linear pair property to find the angle adjacent to \(\small \angle BEC\).
Use vertically opposite angles to identify equal angles at point \(\small E\).
Apply the same segment theorem to relate \(\small \angle ECD\) and \(\small \angle EBA\).
Use the angle sum property in \(\small \triangle ABE\) to find \(\small \angle BAC\).
📊 Graph / Figure Graph / Figure ›
✏️ Solution Complete Solution ›
- Given
- \[\small \angle BEC = 130^\circ\]
- \angle ECD = 20^\circ
- To Find
- \[\small \angle BAC\]
- Step-by-step Solution
- Since lines \(\small AC\) and \(\small BD\) intersect at point \(\small E\), adjacent angles form a linear pair.
- Therefore,\[\small \angle BEC + \angle CED = 180^\circ\]
- Substituting the given value:\[\small 130^\circ + \angle CED = 180^\circ\]
- Subtracting \(\small 130^\circ\) from both sides:\[\small \angle CED = 180^\circ - 130^\circ\]
- \[\small \angle CED = 50^\circ\]
- Now, vertically opposite angles are equal
- Therefore,\[\small \begin{aligned}\angle AEB &= \angle CED\\&= 50^\circ\end{aligned}\]
- Also,\[\small \angle ECD = 20^\circ\]
- Since \(\small E\) lies on line \(\small AC\), \(\small \angle ECD = \angle ACD\).
- Both \(\small \angle ACD\) and \(\small \angle ABD\) subtend the same chord \(\small AD\).
- Therefore, by the theorem:
- Angles in the same segment are equal.\[\small \angle ABD = \angle ACD\]
- \[\small \angle ABE = 20^\circ\]
- Now consider \(\small \triangle ABE\).
- By the angle sum property of a triangle:\[\small \angle BAE + \angle ABE + \angle AEB = 180^\circ\]
- Substituting the known values:\[\small \angle BAE + 20^\circ + 50^\circ = 180^\circ\]
- Adding the known angles:\[\small \angle BAE + 70^\circ = 180^\circ\]
- Subtracting \(\small 70^\circ\) from both sides:\[\small \begin{aligned}\angle BAE &= 180^\circ - 70^\circ\\\angle BAE &= 110^\circ\end{aligned}\]
- Since point \(\small E\) lies on line \(\small AC\),\[\small \angle BAE = \angle BAC\]
- Therefore,\[\small \angle BAC = 110^\circ\]
🎯 Exam Significance Exam Significance ›
This question is highly important because it integrates multiple geometry concepts into a single problem.
Students learn how to:
- Handle intersecting chords inside a circle
- Apply vertically opposite angle properties
- Use same segment theorem correctly
- Apply triangle angle properties systematically
Such integrated geometry problems are frequently asked in:
- CBSE Board Examinations
- NTSE examinations
- Olympiad geometry problems
- JEE Foundation preparation
A common mistake is assuming that the vertically opposite angle at \(\small E\) is \(\small 130^\circ\). Actually, the required angle inside the triangle is the supplementary angle \(\small 50^\circ\).
🔑 Key Takeaways Key Takeaways ›
-
Adjacent angles on a straight line add up to \(\small 180^\circ\).
-
Vertically opposite angles are equal.
-
Angles in the same segment of a circle are equal.
-
Sum of angles in a triangle is \(\small 180^\circ\).
-
Always identify whether the angle needed is supplementary or vertically opposite.
📘 Concept & Theory Theory and Concept ›
This problem uses several important properties of cyclic quadrilaterals and circles.
The main concepts involved are:
- Opposite angles of a cyclic quadrilateral are supplementary.
- Angles subtended by the same chord in the same segment are equal.
- Angles opposite equal sides in an isosceles triangle are equal.
- Sum of angles in a triangle is \(\small 180^\circ\).
Careful identification of common chords is the key to solving this problem correctly.
🗺️ Solution Roadmap Step-by-step Plan ›
Use the same segment theorem to find \(\small \angle BDC\) and \(\small \angle DAC\).
Calculate \(\small \angle BAD\).
Apply the cyclic quadrilateral property to find \(\small \angle BCD\).
Use the condition \(\small AB = BC\) to identify an isosceles triangle.
Use angle sum property in \(\small \triangle BCD\) to find \(\small \angle ECD\).
📊 Graph / Figure Graph / Figure ›
✏️ Solution Complete Solution ›
- Given
\[\small \angle DBC = 70^\circ \]
\[\small \angle BAC = 30^\circ \]
\(\small ABCD\) is a cyclic quadrilateral and diagonals \(\small AC\) and \(\small BD\) intersect at point \(\small E\).
- To Find
\[\small \angle BCD \]
and if \(\small AB = BC\), find:
\[\small \angle ECD \]
- Step-by-step Solution
- First, observe that:
- \(\small \angle BAC\) and \(\small \angle BDC\) subtend the same chord \(\small BC\).
- Therefore, by the same segment theorem:\[\small \begin{aligned}\angle BDC &= \angle BAC\\\angle BDC &= 30^\circ\end{aligned}\]
- Similarly,
- \(\small \angle DBC\) and \(\small \angle DAC\) subtend the same chord \(\small DC\).
- Therefore,\[\small \begin{aligned}\angle DAC &= \angle DBC\\&= 70^\circ\end{aligned}\]
- Now,\[\small \angle BAD = \angle BAC + \angle CAD\]
- Substituting the values:\[\small \begin{aligned}\angle BAD &= 30^\circ + 70^\circ\\&=100^\circ\end{aligned}\]
- Since \(\small ABCD\) is a cyclic quadrilateral, opposite angles are supplementary.
- Therefore,\[\small \angle BAD + \angle BCD = 180^\circ\]
- Substituting \(\small \angle BAD = 100^\circ\): \[\small 100^\circ + \angle BCD = 180^\circ\]
- Subtracting \(\small 100^\circ\) from both sides: \[\small \angle BCD = 180^\circ - 100^\circ\]
- Now suppose:\[\small AB = BC\]
- Therefore, \(\small \triangle ABC\) is an isosceles triangle.
- Angles opposite equal sides are equal.
- Hence,\[\small angle BAC = \angle BCA\]
- Since:\[\small \angle BAC = 30^\circ\]
- Therefore,\[\small \angle BCA = 30^\circ\]
- Now consider \(\small \triangle BCD\).
- We know:\[\small \angle DBC = 70^\circ\]
- \[\small \angle BCD = 80^\circ\]
- Since \(\small E\) lies on diagonal \(\small AC\),\[\small \angle ECD = \angle ACD\]
- Also,\[\small \angle BCD = \angle BCA + \angle ACD\]
- Substituting the known values:\[\small 80^\circ = 30^\circ + \angle ACD\]
- Subtracting \(\small 30^\circ\) from both sides:\[\small \angle ACD = 50^\circ\]
- Therefore,\[\small \angle ECD = 50^\circ\]
🎯 Exam Significance Exam Significance ›
This is a very important multi-concept geometry problem.
Students learn how to combine:
- Same segment theorem
- Properties of cyclic quadrilaterals
- Isosceles triangle properties
- Triangle angle relationships
Such integrated reasoning problems are frequently asked in:
- CBSE Board Examinations
- NTSE examinations
- Olympiad geometry problems
- JEE Foundation preparation
A common mistake is directly using \(\small \angle BCD = 80^\circ\) as \(\small \angle ECD\). Students must remember that:
\[\small \angle BCD = \angle BCA + \angle ACD \]
🔑 Key Takeaways Key Takeaways ›
-
Opposite angles of a cyclic quadrilateral are supplementary.
-
Angles in the same segment of a circle are equal.
-
Equal sides in a triangle imply equal opposite angles.
-
Break complex angles into smaller known angles whenever possible.
-
Careful diagram observation is essential in geometry problems.
📘 Concept & Theory Theory and Concept Used ›
This proof is based on one of the most important theorems in circle geometry:
The angle subtended by a diameter at the circumference is always a right angle.
This theorem is also known as:
\[\small \text{Angle in a semicircle} = 90^\circ \]
Since the diagonals \(\small AC\) and \(\small BD\) are diameters, every angle of the quadrilateral subtended by these diameters becomes a right angle.
A quadrilateral having all four angles equal to \(\small 90^\circ\) is a rectangle.
🗺️ Solution Roadmap Step-by-step Plan ›
Observe that diagonals \(\small AC\) and \(\small BD\) are diameters.
Apply the theorem:
Angle subtended by a diameter at the circumference is \(\small 90^\circ\).Show that all four interior angles of quadrilateral \(\small ABCD\) are right angles.
Conclude that the quadrilateral is a rectangle.
📊 Graph / Figure Graph / Figure ›
📐 Proof Proof ›
- Since \(\small AC\) is a diameter of the circle, the angle subtended by \(\small AC\) at the circumference is a right angle.
- Therefore,\[\small \angle ABC = 90^\circ\]
- Similarly, angle subtended by diameter \(\small AC\) at point \(\small D\) is also a right angle.
- Therefore,\[\small \angle ADC = 90^\circ\]
- Now consider diameter \(\small BD\).
- The angle subtended by diameter \(\small BD\) at point \(\small A\) is a right angle.
- Therefore,\[\small \angle BAD = 90^\circ\]
- Similarly, the angle subtended by diameter \(\small BD\) at point \(\small C\) is also a right angle.
- Therefore,\[\small \angle BCD = 90^\circ\]
- Thus, all four angles of quadrilateral \(\small ABCD\) are right angles.\[\small \angle ABC = \angle BCD = \angle CDA = \angle DAB = 90^\circ\]
- A quadrilateral whose all interior angles are right angles is called a rectangle.
- Therefore,\[\small ABCD \text{ is a rectangle}\]
🎯 Exam Significance Exam Significance ›
This theorem-based proof is extremely important in geometry because it connects:
- Cyclic quadrilaterals
- Diameters of circles
- Angles in semicircles
- Properties of rectangles
Such proof-based questions are frequently asked in:
- CBSE Board Examinations
- School annual examinations
- NTSE examinations
- Olympiad geometry proofs
Students should clearly mention the theorem “Angle in a semicircle is a right angle” while writing the proof to improve presentation and scoring.
🔑 Key Takeaways Key Takeaways ›
-
Angle subtended by a diameter at the circumference is always \(\small 90^\circ\).
-
A cyclic quadrilateral can contain important special quadrilaterals.
-
If all four angles are right angles, the quadrilateral is a rectangle.
-
Circle theorems are extremely useful in geometric proofs.
-
Writing the theorem explicitly strengthens proof presentation.
📘 Concept & Theory Theory and Concept Used ›
This proof uses properties of:
- Isosceles trapezium
- Parallel lines
- Interior angles on the same side of a transversal
- Cyclic quadrilateral theorem
A very important theorem used here is:
If a pair of opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.
Since the non-parallel sides of the trapezium are equal, the trapezium becomes an isosceles trapezium.
In an isosceles trapezium, angles adjacent to the same base are equal.
🗺️ Solution Roadmap Step-by-step Plan ›
Use the fact that \(\small AB \parallel CD\).
Use the property of an isosceles trapezium:
Angles adjacent to the same base are equal.Show that a pair of opposite angles are supplementary.
Apply the converse of the cyclic quadrilateral theorem.
📊 Graph / Figure Graph / Figure ›
📐 Proof Proof ›
- Since:\[\small AD = BC\]
- the trapezium \(\small ABCD\) is an isosceles trapezium.
- In an isosceles trapezium, angles adjacent to the same base are equal.
- Therefore,\[\small \angle A = \angle B\]
- Also, since \(\small AB \parallel CD\), line \(\small BC\) acts as a transversal.
- Hence, interior angles on the same side of the transversal are supplementary.
- Therefore,\[\small \angle B + \angle C = 180^\circ\]
- But:\[\small \angle A = \angle B\]
- Substituting:\[\small \angle A + \angle C = 180^\circ\]
- Thus, one pair of opposite angles of quadrilateral \(\small ABCD\) is supplementary.
- By the converse of the cyclic quadrilateral theorem:
- If a pair of opposite angles of a quadrilateral are supplementary, then the quadrilateral is cyclic.
- Therefore,\[\small ABCD \text{ is a cyclic quadrilateral}\]
🎯 Exam Significance Exam Significance ›
This proof is highly important because it combines:
- Properties of trapeziums
- Parallel line angle properties
- Isosceles trapezium properties
- Converse of cyclic quadrilateral theorem
Such proof-based geometry questions are frequently asked in:
- CBSE Board Examinations
- School annual examinations
- NTSE examinations
- Olympiad geometry proofs
Students often make proofs unnecessarily lengthy. The shortest and most elegant approach is to show:
\[\small \angle A + \angle C = 180^\circ \]
🔑 Key Takeaways Key Takeaways ›
-
Equal non-parallel sides make a trapezium isosceles.
-
Angles adjacent to the same base are equal in an isosceles trapezium.
-
Interior angles on the same side of a transversal are supplementary.
-
A quadrilateral is cyclic if one pair of opposite angles is supplementary.
-
Concise logical proofs score better in board examinations.
📘 Concept & Theory Theory and Concept Used ›
This proof is based on two important geometry concepts:
- Angles in the same segment of a circle are equal.
- Vertically opposite angles are equal.
In a circle:
Angles subtended by the same chord in the same segment are equal.
Also, when two straight lines intersect:
Vertically opposite angles are equal.
This proof becomes very elegant when the equal angles are linked step-by-step.
🗺️ Solution Roadmap Step-by-step Plan ›
Use the same segment theorem in the first circle to relate \(\small \angle ACP\) and \(\small \angle ABP\).
Use vertically opposite angles at point \(\small B\).
Use the same segment theorem in the second circle to relate \(\small \angle DBQ\) and \(\small \angle QCD\).
Connect all equal angles to prove the result.
📊 Graph / Figure Graph / Figure ›
📐 Proof Proof ›
Two circles intersect at points \(\small B\) and \(\small C\).
Through point \(\small B\), line segments \(\small ABD\) and \(\small PBQ\) are drawn, intersecting the circles at points \(\small A, D\) and \(\small P, Q\) respectively.
- First, consider the first circle containing points \(\small A, B, C,\) and \(\small P\).
- Here:
- \(\small \angle ACP\) subtends chord \(\small AP\)
- \(\small \angle ABP\) also subtends the same chord \(\small AP\)
- Therefore, by the same segment theorem:\[\small \angle ACP = \angle ABP\]
- Now consider the intersecting lines at point \(\small B\).
- Angles \(\small \angle ABP\) and \(\small \angle DBQ\) are vertically opposite angles.
- Therefore,\[\small \angle ABP = \angle DBQ\]
- Next, consider the second circle containing points \(\small D, B, C,\) and \(\small Q\).
- Here:
- \(\small \angle DBQ\) subtends chord \(\small DQ\)
- \(\small \angle QCD\) also subtends the same chord \(\small DQ\)
- Therefore, by the same segment theorem:\[\small \angle DBQ = \angle QCD\]
- From:\[\small \angle ACP = \angle ABP\] \[\small \angle ABP = \angle DBQ\] \[\small \angle DBQ = \angle QCD\]
- Therefore,\[\small \angle ACP = \angle QCD\]
- Hence Proved
🎯 Exam Significance Exam Significance ›
This proof is highly important because it develops advanced geometric reasoning using multiple circle theorems together.
Students learn how to:
- Apply the same segment theorem in different circles
- Use vertically opposite angles strategically
- Link equal angles systematically
- Write elegant geometry proofs
Such theorem-based proofs are frequently asked in:
- CBSE Board Examinations
- NTSE examinations
- Olympiad geometry proofs
- JEE Foundation geometry preparation
Students often miss the key observation that the proof uses two different circles separately.
🔑 Key Takeaways Key Takeaways ›
-
Angles subtended by the same chord in the same segment are equal.
-
Vertically opposite angles are always equal.
-
One proof may involve multiple circles simultaneously.
-
Stepwise logical linking is important in geometry proofs.
-
Proper theorem identification improves proof presentation.
📘 Concept & Theory Theory and Concept Used ›
This proof is based on one of the most important circle theorems:
The angle subtended by a diameter at the circumference is always a right angle.
This theorem is also written as:
\[\small \text{Angle in a semicircle} = 90^\circ \]
Since point \(\small M\) lies on both circles:
- \(\small \angle AMB\) becomes a right angle in the first circle.
- \(\small \angle AMC\) becomes a right angle in the second circle.
These two right angles together form a straight angle, proving that points \(\small B\), \(\small M\), and \(\small C\) are collinear.
🗺️ Solution Roadmap Step-by-step Plan ›
Draw two circles with diameters \(\small AB\) and \(\small AC\).
Let the circles intersect at point \(\small M\).
Apply the angle in a semicircle theorem in both circles.
Show that:
\[\small \angle BMC = 180^\circ \]Conclude that \(\small B\), \(\small M\), and \(\small C\) are collinear.
📊 Graph / Figure Graph / Figure ›
📐 Proof Proof ›
- One circle with diameter \(\small AB\)
- Another circle with diameter \(\small AC\)
- Let these two circles intersect at point \(\small M\) other than point \(\small A\).
- We have to prove that point \(\small M\) lies on side \(\small BC\).
- Since \(\small AB\) is the diameter of the first circle, by the theorem:\[\small \text{Angle in a semicircle} = 90^\circ\]
- Therefore,\[\small \angle AMB = 90^\circ\]
- Similarly, \(\small AC\) is the diameter of the second circle.
- Again using the same theorem:\[\small \angle AMC = 90^\circ\]
- Now observe that:\[\small \angle AMB + \angle AMC=90^\circ + 90^\circ\]
- \[\small \angle AMB + \angle AMC = 180^\circ\]
- This means that the angle formed by points \(\small B\), \(\small M\), and \(\small C\) is a straight angle.
- Therefore, points \(\small B\), \(\small M\), and \(\small C\) are collinear.
- Hence, point \(\small M\) lies on side \(\small BC\).
- Therefore,\[\small \text{The point of intersection of the circles lies on the third side } BC\]
🎯 Exam Significance Exam Significance ›
This proof is highly important because it demonstrates how circle theorems can establish collinearity in geometry.
Students learn how to:
- Apply the angle in semicircle theorem
- Use right angles strategically
- Prove collinearity geometrically
- Connect circles with triangle geometry
Such theorem-based proofs are frequently asked in:
- CBSE Board Examinations
- NTSE examinations
- Olympiad geometry problems
- JEE Foundation preparation
Students should clearly state why \(\small \angle AMB = 90^\circ\) and \(\small \angle AMC = 90^\circ\) instead of directly writing the result.
🔑 Key Takeaways Key Takeaways ›
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Angle subtended by a diameter is always a right angle.
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Two right angles together form a straight angle.
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If an angle equals \(\small 180^\circ\), the points are collinear.
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Circle geometry can help prove collinearity elegantly.
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Writing theorem names improves proof presentation.
📘 Concept & Theory Theory and Concept Used ›
This proof is based on two important circle geometry theorems:
- Angle in a semicircle is a right angle.
- Angles subtended by the same chord in the same segment are equal.
Since both triangles \(\small ABC\) and \(\small ADC\) are right triangles with common hypotenuse \(\small AC\):
\[\small \angle ABC = 90^\circ \]
\[\small \angle ADC = 90^\circ \]
Therefore, points \(\small A\), \(\small B\), \(\small C\), and \(\small D\) lie on the same circle having diameter \(\small AC\).
Once the quadrilateral becomes cyclic, equal angles subtended by the same chord can be used.
🗺️ Solution Roadmap Step-by-step Plan ›
Use the fact that both triangles are right triangles.
Apply the theorem:
Angle in a semicircle is \(\small 90^\circ\).Show that points \(\small A\), \(\small B\), \(\small C\), and \(\small D\) lie on the same circle.
Identify the common chord \(\small CD\).
\[\small \angle CAD = \angle CBD \]
📊 Graph / Figure Graph / Figure ›
📐 Proof Proof ›
- \(\small \triangle ABC\) is a right triangle.
- \(\small \triangle ADC\) is a right triangle.
- Both triangles have common hypotenuse \(\small AC\).
Therefore,
\[\small \angle ABC = 90^\circ \]
\[\small \angle ADC = 90^\circ \]
- Since \(\small \angle ABC = 90^\circ\), point \(\small B\) lies on the circle having diameter \(\small AC\).
- Similarly, since \(\small \angle ADC = 90^\circ\), point \(\small D\) also lies on the circle having diameter \(\small AC\).
- Therefore, points \(\small A\), \(\small B\), \(\small C\), and \(\small D\) lie on the same circle.
- Hence, quadrilateral \(\small ABCD\) is cyclic.
- Now consider:
- \(\small \angle CAD\)
- \(\small \angle CBD\)
- Both these angles subtend the same chord \(\small CD\).
- Therefore, by the theorem:
- Angles subtended by the same chord in the same segment are equal.\[\small \angle CAD = \angle CBD\]
- Therefore,\[\small \angle CAD = \angle CBD\]
🎯 Exam Significance Exam Significance ›
- Right triangle properties
- Angle in a semicircle theorem
- Cyclic quadrilateral concepts
- Same segment theorem
Such proof-based geometry questions are frequently asked in:
- CBSE Board Examinations
- NTSE examinations
- Olympiad geometry proofs
- JEE Foundation preparation
Students should clearly explain why \(\small ABCD\) becomes cyclic before applying the same segment theorem.
🔑 Key Takeaways Key Takeaways ›
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Angle in a semicircle is always \(\small 90^\circ\).
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Points subtending a right angle lie on the circle with the hypotenuse as diameter.
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A quadrilateral becomes cyclic if all four points lie on the same circle.
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Angles subtended by the same chord are equal.
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Proper theorem sequencing is essential in geometry proofs.
📘 Concept & Theory Theory and Concept Used ›
This proof combines properties of:
- Cyclic quadrilaterals
- Parallelograms
- Rectangles
The two key theorems used are:
Opposite angles of a cyclic quadrilateral are supplementary.
\[\small \angle A + \angle C = 180^\circ \]
Opposite angles of a parallelogram are equal.
\[\small \angle A = \angle C \]
When an angle is both equal and supplementary to another angle, each angle must be \(\small 90^\circ\).
🗺️ Solution Roadmap Step-by-step Plan ›
Use the cyclic quadrilateral property:
Opposite angles are supplementary.Use the parallelogram property:
Opposite angles are equal.Combine both relations to find each angle.
Show that every angle equals \(\small 90^\circ\).
Conclude that the parallelogram is a rectangle.
📊 Graph / Figure Graph / Figure ›
📐 Proof Proof ›
- Since \(\small ABCD\) is a cyclic quadrilateral, opposite angles are supplementary.
- Therefore,\[\small \angle A + \angle C = 180^\circ\]
- Also, in a parallelogram, opposite angles are equal.
- Therefore,\[\small \angle A = \angle C\]
- Substituting:\[\small \begin{aligned}\angle A + \angle A &= 180^\circ\\&= 180^\circ\\&=90^\circ\end{aligned}\]
- Since opposite angles are equal:\[\small \angle C = 90^\circ\]
- Also, adjacent angles in a parallelogram are supplementary.
- Therefore,\[\small \angle A + \angle B = 180^\circ\]
- Substituting \(\small \angle A = 90^\circ\):\[\small 90^\circ + \angle B = 180^\circ\]
- Subtracting \(\small 90^\circ\) from both sides:\[\small \angle B = 90^\circ\]
- Similarly,\[\small \angle D = 90^\circ\]
- Hence, all four angles of parallelogram \(\small ABCD\) are right angles.\[\small \angle A = \angle B = \angle C = \angle D = 90^\circ\]
- A parallelogram whose all angles are right angles is called a rectangle.
- Therefore,\[\small ABCD \text{ is a rectangle}\]
- Hence Proved
🎯 Exam Significance Exam Significance ›
This proof is highly important because it connects two major geometry concepts:
- Cyclic quadrilateral properties
- Parallelogram properties
Students learn how to combine multiple theorems logically to identify a special quadrilateral.
Such proof-based problems are frequently asked in:
- CBSE Board Examinations
- School annual examinations
- NTSE examinations
- Olympiad geometry proofs
Students should explicitly mention:
\[\small \angle A = \angle C \]
before using:
\[\small \angle A + \angle C = 180^\circ \]
because this logical connection is the heart of the proof.
🔑 Key Takeaways Key Takeaways ›
-
Opposite angles of a cyclic quadrilateral are supplementary.
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Opposite angles of a parallelogram are equal.
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Equal supplementary angles must each be \(\small 90^\circ\).
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A parallelogram with all right angles is a rectangle.
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Combining different geometry properties often leads to elegant proofs.