NCERT Class 9 Maths Exercise 7.2 Solutions

📘 Concept & Theory Concept Used ›

In an isosceles triangle, angles opposite equal sides are equal.
The angle bisector divides an angle into two equal parts.
If two angles of a triangle are equal, then the sides opposite to them are equal.
SAS Congruence Rule: If two sides and the included angle of one triangle are equal to the corresponding two sides and included angle of another triangle, then the triangles are congruent.
CPCT: Corresponding Parts of Congruent Triangles are Equal.

🗺️ Solution Roadmap Step-by-step Plan ›

Use angle bisector property to prove \(\small \angle OBC = \angle OCB \).
From equal angles in \(\small \triangle BOC \), prove \(\small OB = OC \).
Compare triangles \(\small \triangle ABO \) and \(\small \triangle ACO \).
Apply SAS Congruence Rule.
Use CPCT to prove that \(\small AO \) bisects \(\small \angle A \).

📊 Graph / Figure Graph / Figure ›

Fig. 7.29

✏️ Solution Complete Solution ›

Step-by-step Solution · 32 steps

Given
Given that \(\small \triangle ABC \) is an isosceles triangle such that
\[\small AB = AC \]
Also, \(\small BO\) and \(\small CO\) are the bisectors of \(\small \angle B\) and \(\small \angle C\) respectively.
To Prove (i) \(\small OB = OC\)
(ii) \(\small AO\) bisects \(\small \angle A\)
(i) Proof
Find equal angles in triangle \(\small ABC\)
Since \(\small AB = AC\), therefore angles opposite to equal sides are equal.
\[\small \angle B = \angle C\]
Since \(\small BO\) bisects \(\small \angle B\),
\[\small \angle OBC = \frac{1}{2}\angle B\]
Since \(\small CO\) bisects \(\small \angle C\),
\[\small \angle OCB = \frac{1}{2}\angle C\]
But, \[\small \angle B = \angle C\]
Therefore,\[\small \angle OBC = \angle OCB\]
In \(\small \triangle BOC \), we have
\[\small \angle OBC = \angle OCB\]
Sides opposite equal angles are equal.
Therefore, \[\small OB = OC\]
Hence proved
Compare triangles \(\small \triangle ABO \) and \(\small \triangle ACO \)
In triangles \(\small \triangle ABO \) and \(\small \triangle ACO \),
\[\small AB = AC\quad\text{Given}\]
\[\small OB = OC\quad\text{Proved above}\]
also, \[\small \angle ABO = \angle ACO\]
because, \[\small \angle ABO = \frac{1}{2}\angle B\]
\[\small \angle ACO = \frac{1}{2}\angle C\] and \[\small \angle B = \angle C\]
Therefore, \[\small \angle ABO = \angle ACO\]
Hence, by SAS Congruence Rule
\[\small \triangle ABO \cong \triangle ACO\]
Use CPCT
From congruent triangles,
\[\small \angle BAO = \angle CAO\]
Therefore, \(\small AO\) divides \(\small \angle A\) into two equal parts.
Hence,\[\small AO \text{ bisects } \angle A\]

🎯 Exam Significance Exam Significance ›

This question strengthens the concept of isosceles triangles and angle bisectors.
Very important for CBSE board examinations because it combines: congruence, CPCT, and angle properties in a single proof.
Frequently asked in school exams as a proof-based long answer question.
Useful for preparation of Olympiads, NTSE, and other competitive entrance exams where logical geometry proofs are important.
Builds foundation for advanced geometry concepts in higher classes.

📘 Concept & Theory Concept Used ›

A perpendicular bisector of a line segment divides the segment into two equal parts and forms right angles at the point of division.
If two triangles have:
two equal sides and the included angle equal, then the triangles are congruent by SAS Rule.
By CPCT, corresponding sides of congruent triangles are equal.
If two sides of a triangle are equal, then the triangle is an isosceles triangle.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the definition of perpendicular bisector.
Compare triangles \(\small \triangle ABD \) and \(\small \triangle ACD \).
Apply SAS Congruence Rule.
Use CPCT to prove \(AB = AC\).
Conclude that \(\small \triangle ABC \) is isosceles.

📊 Graph / Figure Graph / Figure ›

Fig. 7.30

✏️ Solution Complete Solution ›

Step-by-step Solution · 15 steps

Given
\(\small AD\) is the perpendicular bisector of \(BC\).

Therefore,
\[\small BD = DC \]
and
\[\small \angle ADB = \angle ADC = 90^\circ \]
To Prove\[\small AB = AC\]
Proof
Consider triangles \(\small \triangle ABD \) and \(\small \triangle ACD \)
In triangles \(\small \triangle ABD \) and \(\small \triangle ACD \),
\[\small \angle ADB = \angle ADC = 90^\circ\]
because \(\small AD\) is perpendicular to \(\small BC\).
Also, \[\small BD=DC\]
because \(\small AD\) is the perpendicular bisector of \(\small BC\).
Further, \[\small AD=AD\quad\text{common Side}\]
We have: \[\small \begin{aligned} AD &= AD \\ BD &= DC \\ \angle ADB &= \angle ADC \end{aligned} \]
Therefore, by SAS Congruence Rule,\[\small \triangle ABD \cong \triangle ACD\]
Since corresponding parts of congruent triangles are equal, \[\small AB = AC\quad\text{By CPCT}\]
Hence, \(\small \triangle ABC \) has two equal sides.
Therefore, \[\small \triangle ABC \text{ is an isosceles triangle}\]

🎯 Exam Significance Exam Significance ›

This question is a classic application of perpendicular bisector properties.
Important for CBSE board examinations because it combines: congruence, perpendicular lines, and isosceles triangle properties.
Frequently asked in school examinations as a proof-based geometry question.
Helpful for Olympiads, NTSE, and other competitive entrance exams where geometric reasoning is tested.
Builds strong understanding of triangle congruence and geometric proofs.

📘 Concept & Theory Concept Used ›

In an isosceles triangle, angles opposite equal sides are equal.
An altitude is perpendicular to the side on which it is drawn.
If two angles and one corresponding side of two triangles are equal, then the triangles are congruent by AAS Rule.
By CPCT, corresponding parts of congruent triangles are equal.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the property of isosceles triangle to prove \(\small \angle B = \angle C \).
Observe that \(\small BE\) and \(\small CF\) are altitudes, therefore both form right angles.
Compare triangles \(\small \triangle BFC \) and \(\small \triangle CEB \).
Apply AAS Congruence Rule.
Use CPCT to prove \(\small BE = CF\).

📊 Graph / Figure Graph / Figure ›

Fig. 7.31

✏️ Solution Complete Solution ›

Step-by-step Solution · 18 steps

Find equal angles in the isosceles triangle
Since, \[\small AB = AC\]
therefore angles opposite equal sides are equal.\[\small \angle B = \angle C\]
Hence, \[\small \angle ABC = \angle BCA\]
Observe right angles formed by altitudes
Since \(\small BE\) is an altitude on \(\small AC\),\[\small \angle BEC = 90^\circ\]
Since \(CF\) is an altitude on \(AB\),\[\small \angle CFB = 90^\circ\]
Therefore,\[\small \angle BEC = \angle CFB\]
Compare triangles \(\small \triangle BFC \) and \(\small \triangle CEB \)
In triangles \(\small \triangle BFC \) and \(\small \triangle CEB \),
\[\small BC = CB\quad\text{Common Side}\]
\[\small \angle CFB = \angle BEC\quad \text{ (Each equal to }\(90^\circ\))\]
\angle BCF = \angle CBE
Therefore, by AAS Congruence Rule,\[\small \triangle BFC \cong \triangle CEB\]
Therefore,\[\small BE = CF\quad\text{By CPCT}\]

🎯 Exam Significance Exam Significance ›

This is a very important theorem-based proof related to isosceles triangles.
Frequently asked in CBSE board examinations as a geometry proof question.
Strengthens concepts of: altitudes, congruent triangles, and angle properties.
Helpful for Olympiads, NTSE, and other competitive entrance examinations where logical geometry proofs are tested.
Develops analytical reasoning and theorem application skills.

📘 Concept & Theory Concept Used ›

An altitude of a triangle is perpendicular to the side on which it is drawn.
If two angles and one corresponding side of two triangles are equal, then the triangles are congruent by AAS Rule.
Corresponding Parts of Congruent Triangles are Equal (CPCT).
If two sides of a triangle are equal, then the triangle is an isosceles triangle.

🗺️ Solution Roadmap Step-by-step Plan ›

Observe that \(\small BE\) and \(\small CF\) are altitudes, therefore right angles are formed.
Compare triangles \(\small \triangle ABE \) and \(\small \triangle ACF \).
Apply AAS Congruence Rule.
Use CPCT to prove \(\small AB = AC\)
Conclude that \(\small \triangle ABC \) is an isosceles triangle.

📊 Graph / Figure Graph / Figure ›

Fig. 7.32

✏️ Solution Complete Solution ›

Step-by-step Solution · 12 steps

Given \[\small BE = CF \]
Also, \(\small BE\) and \(\small CF\) are altitudes drawn on \(\small AC\) and \(\small AB\) respectively.

Therefore,
\[\small \angle AEB = \angle AFC = 90^\circ \]
To Prove \[\small \begin{aligned} (i)\;& \triangle ABE \cong \triangle ACF \\ (ii)\;& AB = AC \end{aligned} \]
Proof
Compare triangles \(\small \triangle ABE \) and \(\small \triangle ACF \)
In triangles \(\small \triangle ABE \) and \(\small \triangle ACF \),
\[\small BE = CF\quad\text{Given}\]
\[\small \angle AEB = \angle AFC = 90^\circ\] because \(\small BE\) and \(\small CF\) are altitudes.
Also,\[\small \angle BAE = \angle CAF\] because both represent \(\small \angle A \) of triangle \(\small ABC\).
We have: \[\small \begin{aligned} BE &= CF \\ \angle AEB &= \angle AFC \\ \angle BAE &= \angle CAF \end{aligned} \]
Therefore, by AAS Congruence Rule,\[\small \triangle ABE \cong \triangle ACF\]Hence proved.
Therefore,\[\small AB = AC\quad\text{By CPCT}\]
Since two sides of \(\small \triangle ABC \) are equal, \[\small \triangle ABC \text{ is an isosceles triangle}\]

🎯 Exam Significance Exam Significance ›

This question is a standard theorem-based proof involving altitudes and congruent triangles.
Frequently asked in CBSE board examinations as a proof-based geometry question.
Strengthens concepts of: altitudes, right angles, congruence rules, and CPCT.
Important for Olympiads, NTSE, and other competitive entrance examinations where logical geometric reasoning is tested.
Helps students develop strong proof-writing skills in geometry.

📘 Concept & Theory Concept Used ›

In an isosceles triangle, angles opposite equal sides are equal.
If equal angles are added to equal angles, then the resulting angles are also equal.
Exterior and combined angle observations are very important in geometry proofs.
Careful angle decomposition helps simplify complex proofs.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the isosceles property in \(\small \triangle ABC \) to find equal angles.
Use the isosceles property in \(\small \triangle DBC \) to find another pair of equal angles.
Express \(\small \angle ABD \) and \(\small \angle ACD \) as sums of smaller angles.
Substitute equal angles.
Conclude that \(\small \angle ABD = \angle ACD \).

📊 Graph / Figure Graph / Figure ›

Fig. 7.33

✏️ Solution Complete Solution ›

Step-by-step Solution · 16 steps

Given
\(\small \triangle ABC \) is an isosceles triangle.
\[\small AB = AC \]
Also, \(\small \triangle DBC \) is an isosceles triangle.
\[\small BD = CD \]
To Prove \[\small \angle ABD = \angle ACD\]
Proof
Use isosceles property in \(\small \triangle ABC \)
Since, \[\small AB=AC\]
therefore angles opposite equal sides are equal. \[\small \angle ABC = \angle ACB\]
Use isosceles property in \(\small \triangle DBC \)
Since, \[\small BD = CD\]
therefore angles opposite equal sides are equal. \[\small \angle CBD = \angle BCD\]
Express larger angles as sums. At vertex \(B\),
\[\small \angle ABD = \angle ABC + \angle CBD\]
At vertex \(C\),\[\small \angle ACD = \angle ACB + \angle BCD\]
Substitute equal angles, From previous steps,\[\small \angle ABC = \angle ACB\] and \[\small \angle CBD = \angle BCD\]
Therefore, \[\small \begin{aligned} \angle ABD &= \angle ABC + \angle CBD \\ &= \angle ACB + \angle BCD \\ &= \angle ACD \end{aligned} \]
Hence, \[\small \angle ABD = \angle ACD\]
Proved

🎯 Exam Significance Exam Significance ›

This question develops strong understanding of angle properties in isosceles triangles.
Frequently asked in CBSE board examinations as a proof-based geometry problem.
Improves skills in angle decomposition and logical mathematical reasoning.
Useful for Olympiads, NTSE, and other competitive entrance examinations.
Helps students learn how to combine multiple triangle properties in a single proof.

📘 Concept & Theory Concept Used ›

In an isosceles triangle, angles opposite equal sides are equal.
The sum of interior angles of a triangle is \(\small 180^\circ\).
If a side is produced, exterior angle concepts and linear pair properties become useful.
Careful angle tracking is essential in geometry proofs.

🗺️ Solution Roadmap Step-by-step Plan ›

Use \(\small AB = AC\) to obtain equal base angles in \(\small \triangle ABC \).
Use \(\small AB = AD\) to obtain equal angles in \(\small \triangle ACD \)
Observe that \(\small B, A, D\) are collinear.
Apply the linear pair property at point \(\small A\).
Find the value of \(\small \angle BCD \).

📊 Graph / Figure Graph / Figure ›

Fig. 7.34

✏️ Solution Complete Solution ›

Step-by-step Solution · 28 steps

Given \[\small AB = AC \]
and side \(\small BA\) is produced to \(\small D\) such that
\[\small AD = AB \]
Therefore,
\[\small AB = AC = AD \]
To Prove\[\small \angle BCD = 90^\circ\]
Proof Use isosceles property in \(\small \triangle ABC \)
Since, \[\small AB = AC\]
therefore angles opposite equal sides are equal.
Let \[\small \angle ABC = \angle ACB = x\]
Therefore, in \(\small \triangle ABC \),
\[\small \angle BAC + x + x = 180^\circ\]
\[\small \angle BAC = 180^\circ - 2x\]
Use isosceles property in \(\small \triangle ACD \)
Since \[\small AC = AD\]
therefore angles opposite equal sides are equal. Hence,\[\small \angle ACD = \angle CDA\]
Let each of them be equal to \(\small y\).
Therefore, in \(\small \triangle ACD \),
\[\small \angle CAD + y + y = 180^\circ\]
\[\small \angle CAD = 180^\circ - 2y\]
Use linear pair property at \(\small A\)
Since \(\small B, A, D\) are collinear,
\[\small \angle BAC + \angle CAD = 180^\circ\]
Substituting the values,
\[\small \begin{aligned}(180^\circ - 2x) + (180^\circ - 2y) &= 180^\circ\\ 360^\circ - 2x - 2y &= 180^\circ\\
2x + 2y &= 180^\circ\\ x + y &= 90^\circ\end{aligned}\]
Find \(\small \angle BCD \)
At point \(C\),\[\small \angle BCD = \angle BCA + \angle ACD\]
Therefore, \[\small \angle BCD = x + y\]
But from Step 3,
\[\small x + y = 90^\circ\]
Hence, \angle BCD = 90^\circ

🎯 Exam Significance Exam Significance ›

This is an important proof involving isosceles triangles and angle relationships.
Frequently asked in CBSE board examinations as a reasoning-based geometry proof.
Strengthens understanding of: linear pairs, angle sum property, and isosceles triangle properties.
Very useful for Olympiads, NTSE, and competitive entrance examinations.
Helps students avoid common mistakes in angle calculations and proof writing.

📘 Concept & Theory Concept Used ›

The sum of all interior angles of a triangle is \(180^\circ\).
In an isosceles triangle, angles opposite equal sides are equal.
A triangle having one angle equal to \(90^\circ\) is called a right angled triangle.
If two equal angles together add up to \(90^\circ\), then each angle measures \(45^\circ\).

🗺️ Solution Roadmap Step-by-step Plan ›

Use the angle sum property of a triangle.
Use the isosceles triangle property \(\small AB = AC\).
Form an equation using equal angles.
Solve step-by-step to find \(\small \angle B \) and \(\small \angle C \).

✏️ Solution Complete Solution ›

Step-by-step Solution · 20 steps

Given
\(\small \triangle ABC \) is a right angled triangle such that
\[\small \angle A = 90^\circ \]
Also,
\[\small AB = AC \]
To Find \[\small \angle B \text{ and } \angle C\]
Proof
Use angle sum property
We know that the sum of interior angles of a triangle is \(\small 180^\circ\).
\[\small \angle A + \angle B + \angle C = 180^\circ\]
Substituting \(\small \angle A = 90^\circ \),
\[\small 90^\circ + \angle B + \angle C = 180^\circ\]
Subtracting \(\small 90^\circ\) from both sides,
\[\angle B + \angle C = 90^\circ\]
Since, \[\small AB=AC\]
therefore angles opposite equal sides are equal.
\[\small \angle B = \angle C\]
Let \[\small \angle B = \angle C = x\]
Substituting in \(\small \angle B + \angle C = 90^\circ \),
\[\small x + x = 90^\circ\]
\[\small 2x = 90^\circ\]
Dividing both sides by \(2\),
\[\small x-45^\circ\]
Therefore,\[small \angle B = 45^\circ\] and \[small \angle C = 45^\circ\]

💡 Answer Final Answer ›

Final Answer: \(\angle B = 45^\circ\)

🎯 Exam Significance Exam Significance ›

2-mark standard board question.

📘 Concept & Theory Concpet Used ›

An equilateral triangle has all three sides equal.
In a triangle, angles opposite equal sides are equal.
The sum of interior angles of a triangle is \(\small 180^\circ\).
If three equal angles together make \(\small 180^\circ\), then each angle is \(\small 60^\circ\).

🗺️ Solution Roadmap Step-by-step Plan ›

Use the definition of an equilateral triangle.
Prove that all three angles are equal.
Apply the angle sum property of a triangle.
Solve step-by-step to find the value of each angle.

✏️ Solution Complete Solution ›

Step-by-step Solution · 18 steps

Let \(\small \triangle ABC \) be an equilateral triangle.
Therefore,\[\small AB = BC = CA\]
To Show:\[\small \angle A = \angle B = \angle C = 60^\circ\]
Use equal side property.
Since,\[\small AB = BC\]
therefore angles opposite equal sides are equal.\[\small \angle C = \angle A\]
Also,\[\small BC = CA\]
therefore,\[\small \angle A = \angle B\]
Hence, \[\small \angle A = \angle B = \angle C\]
Apply angle sum property
Let each angle be equal to \(x\)
\[\small \angle A = \angle B = \angle C = x\]
We know that the sum of interior angles of a triangle is \(180^\circ\).
\[\small \angle A + \angle B + \angle C = 180^\circ\]
Substituting the values, \[\small \begin{aligned}x + x + x &= 180^\circ\\ 3x=180^\circ\end{aligned} \]
Dividing both sides by \(3\),\[\small \begin{aligned}x &= \frac{180^\circ}{3}\\x &= 60^\circ\end{aligned}\]
Therefore,\[\small \angle A = 60^\circ\] \[\small \angle B = 60^\circ\] \[\small \angle C = 60^\circ\]
Hence Proved

🎯 Exam Significance Exam Significance ›

This is one of the most fundamental results related to equilateral triangles.
Frequently asked in CBSE board examinations as a direct proof question.
Builds strong understanding of: angle sum property, isosceles triangle properties, and symmetry in geometry.
Very useful for Olympiads, NTSE, and competitive entrance examinations.
Forms the foundation for advanced geometry and trigonometry concepts.

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