# Ex.8.1 Q9 Quadrilaterals Solution - NCERT Maths Class 9

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## Question

In parallelogram $$ABCD$$, two points $$P$$ and $$Q$$ are taken on diagonal $$BD$$ such that

$$DP = BQ$$ (see the given figure). Show that:

\begin{align}&{\text { (i) } \triangle \mathrm{APD} \cong \Delta \mathrm{CQB}} \\ &{\text { (ii) } \mathrm{AP}=\mathrm{CQ}} \\ &{\text { (iii) } \triangle \mathrm{AQB} \cong \Delta \mathrm{CPD}} \\ &{\text { (iv) } \mathrm{AQ}=\mathrm{CP}} \\ &{\text { (v) APCQ is a parallelogram }}\end{align}

Video Solution
Ex 8.1 | Question 9

## Text Solution

What is known?

$$ABCD$$ is a parallelogram and $$DP = BQ$$

What is unknown?

How we can show that

\begin{align} &\left(\text{i} \right)\text{ }\ \ \ \Delta APD\cong \Delta CQB \\ &\left( \text{ii} \right)\text{ }\ \ AP=CQ \\ &\left( \text{iii} \right)\text{ }\ \Delta AQB\cong \Delta CPD \\ &\left(\text{ iv} \right)\text{ }\ AQ=CP \\ &\left( \text{v} \right)\text{ }\ APCQ\text{ }is\text{ }a\text{ parallelogram} \end{align}

Reasoning:

We can use alternate interior angles and parallelogram property for triangles congruence criterion to show triangles congruent then we can say corresponding parts of congruent triangles will be equal.

Steps:

(i) $$In{\text{ }}\Delta APD\;{\text{and}}{\text{ }}\Delta CQB,$$

\begin{align} \angle \mathrm{ADP} &\!=\!\angle \mathrm{CBQ}\!\begin{bmatrix}\text {Alternate interior}\\ \text{angles for BC } \| \mathrm{AD}\!\end{bmatrix} \\ \mathrm{AD}&\!=\!\mathrm{CB}\begin{bmatrix}\text {Opposite sides of} \\\text{parallelogram } \mathrm{ABCD}\end{bmatrix} \\ \mathrm{DP} &=\mathrm{BQ}\;(\text {Given}) \\ \therefore\triangle \mathrm{APD} &\cong \Delta \mathrm{CQB}\;\begin{bmatrix}\text { Using SAS }\\\text{congruence rule }\end{bmatrix} \end{align}

(ii) As we had observed that

$$\Delta APD \cong \Delta CQB$$

\begin{align} \therefore\, \mathrm{AP}=\mathrm{CQ}(\mathrm{CPCT})\end{align}

(iii) In $$\Delta AQB\;{\text{and}}{\text{ }}\Delta CPD,$$

\begin{align}{\angle \mathrm{ABQ}}&\!=\!\angle \mathrm{CDP}\!\begin{bmatrix}\! \text {Alternate interior}\\\text{ angles for } \mathrm{AB} \| \mathrm{CD} \!\end{bmatrix} \\ {\mathrm{AB}}&\!=\!\mathrm{CD}\begin{bmatrix}\text {Opposite sides of} \\\text{parallelogram } \mathrm{ABCD}\end{bmatrix} \\ {\mathrm{BQ}}&={\mathrm{DP}\;(\text {Given})} \\ {\therefore \Delta \mathrm{AQB} }&\cong \triangle \mathrm{CPD}\;\begin{bmatrix}\text { Using SAS }\\\text{congruence rule }\end{bmatrix} \end{align}

(iv) As we had observed that

$$\Delta AQB \cong \Delta CPD,$$

\begin{align}\therefore \mathrm{AQ}=\mathrm{CP}(\mathrm{CPCT})\end{align}

(v) From the result obtained in (ii) and (iv),

\begin{align} \text { AQ }&=\mathrm{CP} \text { and } \\ {\mathrm{AP}}&={\mathrm{CQ}}\end{align}

Since opposite sides in quadrilateral $$APCQ$$ are equal to each other, $$APCQ$$ is a parallelogram.

Video Solution