NCERT Class 9 Quadrilateral Exercise 8.1 Solutions

📘 Concept & Theory Concept Used ›

In a parallelogram, opposite sides are equal and parallel.
Diagonals of a parallelogram bisect each other.
If the diagonals of a parallelogram are equal, then the parallelogram becomes a rectangle.
To prove a quadrilateral is a rectangle, it is sufficient to prove that one angle is \(90^\circ\).
Congruent triangles can be proved using SSS congruence criterion.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the property that diagonals of a parallelogram bisect each other.
Prove two triangles congruent using SSS criterion.
Show that adjacent angles become equal.
Use the supplementary angle property of a parallelogram.
Conclude that each angle is \(\small 90^\circ\), hence the parallelogram is a rectangle.

📊 Graph / Figure Graph / Figure ›

Fig. 8.10

📐 Proof Proof ›

📌 Given

\(\small ABCD\) is a parallelogram and its diagonals are equal.\[\small AC = BD \]

Let diagonals \(\small AC\) and \(\small BD\) intersect at point \(O\).

🎯 To Prove

\[\small ABCD \text{ is a rectangle}\]

✍️ Proof

Step-by-step Proof · 20 steps

Since \(\small ABCD\) is a parallelogram, opposite sides are equal.
\[\small AB=CD\]
Also, diagonals of a parallelogram bisect each other.
Therefore,\[\small AO = OC \] and \[\small BO = OD \]
\[\small AC = BD\]
Given
Since:\[\small AC = AO + OC \] and \[\small BD = BO + OD \]
But:\[\small AO = OC \] and \[\small BO = OD \]
Hence,\[\small AO = BO \] and \[\small OC = OD \]
Now consider triangles \(\small \triangle AOB \) and \(\small \triangle COD \).
We have:\[\small AO = CO \] \[\small BO = DO \] \[\small AB = CD \]
Therefore,\[\small \triangle AOB \cong \triangle COD\]
by SSS congruence criterion.
Hence corresponding angles are equal.\[\small \angle BAO = \angle DCO\]
Since \(\small AB \parallel CD\) and transversal \(\small AC\) cuts them, alternate interior angles are equal.
Therefore adjacent angles of the parallelogram become equal.
In a parallelogram, adjacent angles are supplementary.\[\small \angle A + \angle B = 180^\circ\]
Also,\[\small \angle A = \angle B\]
Therefore,\[\small\begin{aligned} \angle A + \angle A &= 180^\circ\\ 2\angle A &= 180^\circ\\ \angle A &= 90^\circ \end{aligned}\]
Thus one angle of parallelogram \(\small ABCD\) is \(\small 90^\circ\).
Hence all angles of the parallelogram are \(\small 90^\circ\).
Therefore,\[\small ABCD \text{ is a rectangle}\]

Hence proved

🎯 Exam Significance Exam Significance ›

This question is frequently asked in CBSE Board Examinations as a proof-based theorem question.
It develops understanding of properties of parallelogram and rectangle.
Questions based on diagonals of quadrilaterals are important for NTSE, Olympiads and other competitive entrance examinations.
This problem strengthens concepts of triangle congruence and angle properties.

📘 Concept & Theory Concept Used ›

A square is both a rectangle and a rhombus.
Since a square is a rectangle, its diagonals are equal.
Since a square is a rhombus, its diagonals bisect each other at right angles.
The diagonals of a parallelogram bisect each other.
Pythagoras theorem is used to calculate the length of diagonals.
Congruent triangles can be proved by SSS congruence criterion.

🗺️ Solution Roadmap Step-by-step Plan ›

Use Pythagoras theorem to prove that both diagonals are equal.
Use the property of parallelogram to show diagonals bisect each other.
Prove two triangles congruent using SSS criterion.
Show that the angles formed at the intersection of diagonals are equal.
Conclude that each angle is \(\small 90^\circ\).

📊 Graph / Figure Graph / Figure ›

Fig. 8.10-1

📐 Proof Proof ›

📌 Given

\(\small ABCD\) is a square.

Its diagonals \(\small AC\) and \(\small BD\) intersect at point \(\small O\).

Since all sides of a square are equal,

\[\small AB = BC = CD = DA \]

Also, each angle of a square is \(\small 90^\circ\).

🎯 To Prove

\[\small AC = BD \] \[\small AO = OC \] \[\small BO = OD \]

and diagonals intersect at right angles.

✍️ Proof

Step-by-step Proof · 25 steps

Prove that diagonals are equal
Consider right triangle \(\small \triangle ABC \).
Since angle \(\small B = 90^\circ\), by Pythagoras theorem:\[\small AC^2 = AB^2 + BC^2\]
Let each side of the square be \(a\).
Therefore,\[\small AB = BC = a\]
Substituting: \[\small\begin{aligned} AC^2 &= a^2 + a^2\\AC^2 &= 2a^2\\AC &= \sqrt{2a^2}\\AC &= \sqrt{2}a\end{aligned}\]
Similarly, in right triangle \(\small \triangle BAD \),\[small BD^2 = AB^2 + AD^2\]
Since, \[\small AB = AD = a\]
Therefore,\[\small\begin{aligned} BD^2 &= a^2 + a^2\\BD^2 &= 2a^2\\BD &= \sqrt{2}a \end{aligned}\]
Hence,\[\small AC = BD\]
Step 2: Prove that diagonals bisect each other
A square is also a parallelogram.
The diagonals of a parallelogram bisect each other.
Therefore,\[\small AO = OC \] and \[\small BO = OD \]
Step 3: Prove that diagonals intersect at right angles
Consider triangles \(\small \triangle AOD \) and \(\small \triangle COD \).
We have:\[\small AD = CD\]
because all sides of a square are equal.
Also,\[\small OD=OD\]
(common side)
\[\small AO = OC\]
diagonals bisect each other.
Therefore,\[\small \triangle AOD \cong \triangle COD\]
SSS congruence criterion
Hence corresponding angles are equal.\[\small \angle AOD = \angle COD\]
But points \(\small A\), \(\small O\), and \(\small C\) lie on the same straight line.
Therefore,\[\small \angle AOD + \angle COD = 180^\circ\]
Since both angles are equal,\[\small \begin{aligned}2\angle AOD &= 180^\circ\\\angle AOD &= 90^\circ \end{aligned}\]
Hence the diagonals intersect at right angles.

Therefore, the diagonals of a square are equal, bisect each other, and intersect at right angles.

🎯 Exam Significance Exam Significance ›

This is one of the most important theorem-based questions from quadrilaterals.
Frequently asked in CBSE Board Exams in proof and reasoning format.
Important for Olympiads, NTSE and foundation level entrance examinations.
Helps students understand the relation between square, rectangle and rhombus.
Strengthens understanding of diagonals and congruent triangles.

📘 Concept & Theory Concept Used ›

In a parallelogram, opposite sides are equal.
A diagonal divides a parallelogram into two congruent triangles.
Corresponding parts of congruent triangles are equal (CPCT).
If all four sides of a quadrilateral are equal, then it is a rhombus.
Angle bisector properties are frequently used in quadrilateral proofs.

🗺️ Solution Roadmap Step-by-step Plan ›

Consider triangles \(\small \triangle DAC \) and \(\small \triangle CAB \).
Use opposite sides of a parallelogram and common side property.
Prove triangles congruent using SSS criterion.
Apply CPCT to prove that diagonal \(\small AC\) bisects \(\small \angle C\).
Use equality of sides to prove that \(ABCD\) is a rhombus.

📊 Graph / Figure Graph / Figure ›

Fig. 8.11

📐 Proof Proof ›

📌 Given

\(\small ABCD\) is a parallelogram and diagonal \(\small AC\) bisects \(\small \angle A\).

Therefore, \[\small \angle DAC = \angle CAB \]

Also, in parallelogram \(ABCD\),

\[\small AB = CD \] and \[\small BC = AD \]

🎯 To Prove

(i) \(\small AC\) bisects \(\small \angle C\)

(ii) \(\small ABCD\) is a rhombus

✍️ Proof

Step-by-step Proof · 15 steps

Step 1: Consider triangles \(\small \triangle DAC \) and \(\small \triangle CAB \)
In triangles \(\small \triangle DAC \) and \(\small \triangle CAB \):\[\small AD = BC\]
(Opposite sides of a parallelogram)
\[\small DC = AB\]
(Opposite sides of a parallelogram)
\[\small AC = AC\]
(Common side)
Therefore,\[\small \triangle DAC \cong \triangle CAB\]
by SSS congruence criterion.
Step 2: Prove that \(\small AC\) bisects \(\small \angle C\)
By CPCT,\[\small \angle DCA = \angle BCA\]
Therefore, diagonal \(\small AC\) bisects \(\small \angle C\).
Hence,\[\small AC \text{ bisects } \angle C\]
Step 3: Prove that \(\small ABCD\) is a rhombus
Since diagonal \(\small AC\) bisects \(\small \angle A\),\[\small \angle DAC = \angle CAB\]
Also, from congruent triangles,\[\small AD = AB\]
But in a parallelogram,\[\small AB = CD \] and \[\small AD = BC \]
Therefore,\[\small AB = BC = CD = DA\]
Since all four sides are equal,\[\small ABCD \text{ is a rhombus}\]

Hence Proved

🎯 Exam Significance Exam Significance ›

This is an important proof-based question for CBSE Board Examinations.
Frequently asked in exams to test understanding of congruent triangles.
Important for Olympiads and foundation-level competitive examinations.
Develops logical reasoning and theorem-application skills.
Strengthens concepts of parallelogram, rhombus and angle bisectors.

📘 Concept & Theory Concept Used ›

All angles of a rectangle are \(\small 90^\circ\).
If a right triangle has one acute angle equal to \(\small 45^\circ\), then it is an isosceles right triangle.
In an isosceles triangle, sides opposite equal angles are equal.
A rectangle having all sides equal is a square.
Congruent triangles can be proved using SSS criterion.
By CPCT, corresponding angles of congruent triangles are equal.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the fact that all angles of a rectangle are \(\small 90^\circ\).
Since diagonal \(\small AC\) bisects \(\small \angle A\) and \(\small \angle C\), show that triangles become right isosceles triangles.
Prove all sides of the rectangle are equal.
Conclude that the rectangle is a square.
Use congruent triangles to prove that diagonal \(\small BD\) bisects \(\small \angle B\) and \(\small \angle D\).

📊 Graph / Figure Graph / Figure ›

Fig. 8.11-1

📐 Proof Proof ›

📌 Given

\(\small ABCD\) is a rectangle.

Diagonal \(\small AC\) bisects \(\small \angle A\) and \(\small \angle C\).

Therefore,

\[\small \angle DAC = \angle CAB \] and \[\small \angle BCA = \angle ACD \]

🎯 To Prove

(i) \(\small ABCD\) is a square

(ii) diagonal \(\small BD\) bisects \(\small \angle B\) and \(\small \angle D\)

✍️ Proof

Step-by-step Proof · 25 steps

Part (i): Prove that \(\small ABCD\) is a square
Since \(\small ABCD\) is a rectangle,\[\small \angle A = \angle B = \angle C = \angle D = 90^\circ\]
Diagonal \(\small AC\) bisects \(\small \angle A\).
Given
Therefore,\[\small \angle CAB = \angle DAC\]
But, \[\small \angle A = 90^\circ\]
Hence,\[\small \begin{aligned}\angle CAB = \angle DAC &= \frac{90^\circ}{2}\\\angle CAB = \angle DAC &= 45^\circ \end{aligned}\]
In triangle \(\small \triangle ABC \),\[\small \angle ABC = 90^\circ\] and \[\small \angle CAB = 45^\circ\]
Using angle sum property of triangle:\[\small \begin{aligned}\angle ACB &= 180^\circ - 90^\circ - 45^\circ\\\angle ACB &= 45^\circ \end{aligned}\]
Therefore,\[\small \angle CAB = \angle ACB\]
Hence, sides opposite equal angles are equal.\[\small AB = BC \tag{1}\]
Since opposite sides of a rectangle are equal,\[\small AB = CD\] and \[\small BC = AD\]
Using equation (1),\[\small AB = BC = CD = AD\]
Thus all four sides are equal.
A rectangle with all sides equal is a square.
Therefore,\[\small ABCD \text{ is a square}\]
Part (ii): Prove that diagonal \(BD\) bisects \(\angle B\) and \(\angle D\)
Since \(\small ABCD\) is now proved to be a square,\[\small AB = BC = CD = AD\]
Consider triangles \(\small \triangle ABD \) and \(\small \triangle CBD \).
We have:\[\small \[\small AB = BC \] \[\small AD = CD \] \[\small BD = BD\quad\text{(Common Side)} \]
Therefore,\[\small \triangle ABD \cong \triangle CBD\]
by SSS congruence criterion.
By CPCT,\[\small \angle ABD = \angle DBC\]
Therefore, diagonal \(\small BD\) bisects \(\small \angle B\).
Again, by CPCT,\[\small \angle ADB = \angle BDC\]
Therefore, diagonal \(\small BD\) also bisects \(\small \angle D\).
Hence,\[\small BD \text{ bisects } \angle B \text{ and } \angle D\]

Q.E.D.

🎯 Exam Significance Exam Significance ›

Important theorem-based proof question for CBSE examinations.
Frequently asked in long-answer and reasoning-based questions.
Strengthens concepts of rectangle, square and congruent triangles.
Important for Olympiads, NTSE and foundation-level competitive examinations.
Develops analytical and stepwise proof-writing skills.

Q5

NUMERIC3 marks

In parallelogram \(\small ABCD\), two points \(\small P\) and \(Q\) are taken on diagonal \(\small BD\) such that \(\small DP = BQ\). Show that:
(i) \(\small \triangle APD \cong \triangle CQB\)
(ii) \(\small AP = CQ\)
(iii) \(\small \triangle AQB \cong \triangle CPD\)
(iv) \(\small AQ = CP\)
(v) \(\small APCQ\) is a parallelogram.

📘 Concept & Theory concept Used ›

Opposite sides of a parallelogram are equal and parallel.
Alternate interior angles are equal when a transversal cuts parallel lines.
SAS congruence criterion is used to prove congruent triangles.
By CPCT, corresponding parts of congruent triangles are equal.
If both pairs of opposite sides of a quadrilateral are equal, then the quadrilateral is a parallelogram.

🗺️ Solution Roadmap Step-by-step Plan ›

Use parallel sides of the parallelogram to identify equal alternate angles.
Prove triangles congruent using SAS criterion.
Apply CPCT to obtain equality of corresponding sides.
Repeat the process for another pair of triangles.
Use opposite side equality to prove that quadrilateral \(APCQ\) is a parallelogram.

📊 Graph / Figure Graph / Figure ›

Fig. 8.12

📐 Proof Proof ›

📌 Given

\(\small ABCD\) is a parallelogram.

Points \(\small P\) and \(\small Q\) lie on diagonal \(\small BD\) such that

\[\small DP = BQ \]

Also, in a parallelogram:

\[\small AB = CD \] and \[\small AD = BC \]

🎯 To Prove

(i) \(\small \triangle APD \cong \triangle CQB \)

(ii) \(\small AP = CQ \)

(iii) \(\small \triangle AQB \cong \triangle CPD \)

(iv) \(\small AQ = CP \)

(v) \(\small APCQ \) is a parallelogram

✍️ Proof

Step-by-step Proof · 19 steps

Part (i): Prove \(\small \triangle APD \cong \triangle CQB \)
In triangles \(\small \triangle APD \) and \(\small \triangle CQB \):\[\small DP = BQ\]
Given
Since \(\small AD \parallel BC\) and \(\small BD\) is a transversal, alternate interior angles are equal.\[\small \angle ADP = \angle CBQ\]
Also,\[\small AD = BC\]
(Opposite sides of a parallelogram are equal)
Therefore,\[\small \triangle APD \cong \triangle CQB\]
by SAS congruence criterion.
Part (ii): Prove \(AP = CQ\)
From congruent triangles,\[\small AP = CQ\]
by CPCT.
Part (iii): Prove \( \triangle AQB \cong \triangle CPD \)
In triangles \(\small \triangle AQB \) and \(\small \triangle CPD \)\[\small BQ = DP\]:
(Given)
Since \(\small AB \parallel CD\) and \(\small (BD\) is a transversal, alternate interior angles are equal.\[\small \angle ABQ = \angle CDP\]
Also,\[\small AB = CD\]
(Opposite sides of a parallelogram are equal)
Therefore,\[\small \triangle AQB \cong \triangle CPD\]
by SAS congruence criterion.
Part (iv): Prove \(AQ = CP\)
From congruent triangles,\[\small AQ = CP\]
by CPCT.
Part (v): Prove \(\small APCQ\) is a parallelogram
From parts (ii) and (iv),\[\small AP = CQ \] and \[\small AQ = CP \]
Thus, both pairs of opposite sides of quadrilateral \(\small APCQ\) are equal.
A quadrilateral whose opposite sides are equal is a parallelogram.
Therefore, \[\small APCQ \text{ is a parallelogram } \]

Q.E.D.

🎯 Exam Significance Exam Significance ›

This is an important multi-concept proof question from quadrilaterals.
Frequently asked in CBSE Board Exams for long-answer reasoning practice.
Helps students master congruent triangles and parallelogram properties together.
Important for NTSE, Olympiads and foundation-level competitive examinations.
Strengthens logical proof-writing and theorem-application skills.

📘 Concept & Theory Concept Used ›

Opposite sides of a parallelogram are equal and parallel.
A perpendicular forms a right angle of \(90^\circ\).
Alternate interior angles are equal when a transversal cuts parallel lines.
AAS congruence criterion can be used to prove congruent triangles.
By CPCT, corresponding parts of congruent triangles are equal.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the property that perpendiculars form right angles.
Use parallel sides of the parallelogram to identify equal alternate angles.
Use equality of opposite sides of the parallelogram.
Prove the triangles congruent using AAS criterion.
Apply CPCT to prove \(\small AP = CQ\).

📊 Graph / Figure Graph / Figure ›

Fig. 8.13

📐 Proof Proof ›

📌 Given

\(\small ABCD\) is a parallelogram.

\(\small AP\) and \(\small CQ\) are perpendiculars drawn from vertices \(\small A\) and \(\small C\) respectively on diagonal \(\small BD\).

Therefore,

\[\small AP \perp BD \] and \[\small CQ \perp BD \]

🎯 To Prove

(i) \(\small \triangle APB \cong \triangle CQD\)

(ii) \(\small AP = CQ\)

✍️ Proof

Step-by-step Proof · 19 steps

Step 1: Identify equal sides
Since \(\small ABCD\) is a parallelogram, opposite sides are equal.
Therefore, \[\small AB = CD\]
Step 2: Identify right angles
Since \(\small AP \perp BD\),\[\small \angle APB = 90^\circ\]
Also, since \(CQ \perp BD\),\[\small \angle CQD = 90^\circ\]
Therefore, \[\small \angle APB = \angle CQD\]
Step 3: Identify another pair of equal angles
In parallelogram \(\small ABCD\),\[\small AB \parallel CD\]
Diagonal \(BD\) acts as a transversal.
Therefore alternate interior angles are equal.\[\small \angle PBA = \angle QDC\]
Step 4: Prove congruence
In triangles \(\small \triangle APB \) and \(\small \triangle CQD \): \[\small AB = CD \] \[\small \angle APB = \angle CQD \] \[\small \angle PBA = \angle QDC \]
Therefore,\[\small \triangle APB \cong \triangle CQD\]
by AAS congruence criterion.
Hence,\[\small \triangle APB \cong \triangle CQD\]
Step 5: Prove \(AP = CQ\)
Since corresponding parts of congruent triangles are equal,
\[\small AP = CQ\]
By CPCT
Therefore,\[\small AP = CQ\]

Q.E.D.

🎯 Exam Significance Exam Significance ›

Important proof-based question from properties of parallelogram.
Frequently asked in CBSE examinations and school assessments.
Strengthens concepts of congruent triangles and angle relations.
Useful for Olympiads, NTSE and foundation-level entrance examinations.
Develops stepwise theorem-writing and geometric reasoning skills.

📘 Concept & Theory Concept Used ›

A trapezium having equal non-parallel sides is called an isosceles trapezium.
In an isosceles trapezium, angles adjacent to the same base are equal.
SAS congruence criterion is used to prove congruent triangles.
By CPCT, corresponding parts of congruent triangles are equal.
Diagonals of an isosceles trapezium are equal.

🗺️ Solution Roadmap Step-by-step Plan ›

Identify that the trapezium is an isosceles trapezium because \(AD = BC\).
Use the property of isosceles trapezium to prove equal base angles.
Use SAS criterion to prove triangles congruent.
Apply CPCT to prove equality of diagonals.

📊 Graph / Figure Graph / Figure ›

Fig. 8.14

📐 Proof Proof ›

📌 Given

\[\small AB \parallel CD \] and \[\small AD = BC \]

🎯 To Prove

\[\small \begin{aligned} (i)\ &\angle A = \angle B \\ (ii)\ &\angle C = \angle D \\ (iii)\ &\triangle ABC \cong \triangle BAD \\ (iv)\ &AC = BD \end{aligned} \]

✍️ Proof

Step-by-step Proof · 22 steps

Step 1: Identify the type of trapezium
Since the non-parallel sides of trapezium \(\small ABCD\) are equal,\[\small AD = BC\]
therefore \(\small ABCD\) is an isosceles trapezium.
Part (i): Prove \(\angle A = \angle B\)
In an isosceles trapezium, angles adjacent to the same base are equal.
Therefore,\[\small \angle DAB = \angle CBA\]
Hence,\[\small \angle A = \angle B\]
Part (ii): Prove \(\angle C = \angle D\)
Similarly, in an isosceles trapezium, angles adjacent to the other base are also equal.
Therefore,\[\small \angle ADC = \angle BCD\]
Hence,\[\small \angle C = \angle D\]
Part (iii): Prove \( \triangle ABC \cong \triangle BAD \)
Consider triangles \(\small \triangle ABC \) and \(\small \triangle BAD \).
We have:\[\small AB = BA\]
(Common side)
\[\small BC = AD\]
(Given)
\[\small \angle CBA = \angle DAB\]
(From part (i))
Therefore,\[\small \triangle ABC \cong \triangle BAD\]
by SAS congruence criterion.
Hence,\[\small \triangle ABC \cong \triangle BAD\]
Part (iv): Prove \(AC = BD\)
From congruent triangles \(\small \triangle ABC \) and \(\small \triangle BAD \), corresponding sides are equal.
Therefore,\[\small AC = BD\]
by CPCT.
Hence,\[\small AC = BD\]

Q.E.D.

🎯 Exam Significance Exam Significance ›

Important proof-based question from properties of trapezium and congruent triangles.
Frequently asked in CBSE Board examinations and school tests.
Strengthens understanding of isosceles trapezium properties.
Useful for Olympiads, NTSE and foundation-level competitive examinations.
Develops logical geometry proof-writing and theorem application skills.

QUADRILATERALS

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