# Ex.8.1 Q10 Quadrilaterals Solution - NCERT Maths Class 9

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## Question

$$ABCD$$ is a parallelogram and $$AP$$ and $$CQ$$ are perpendiculars from vertices $$A$$ and $$C$$ on

diagonal $$BD$$ (See the given figure). Show that

(i) \begin{align} &{ \Delta \mathrm{APB} }{\cong \Delta \mathrm{CQD}} \end{align}

(ii) \begin{align} &{ \mathrm{AP}=\mathrm{CQ}}\end{align}

Video Solution
Ex 8.1 | Question 10

## Text Solution

What is known?

$$ABCD$$ is a parallelogram and $$AP\bot DB,\,CQ\bot DB$$

What is unknown?

How we can show that (i) $$\Delta APB\cong \Delta CQD$$

(ii) $$AP = CQ$$

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 \begin{align} \;&\Delta \mathrm{APB} \text { and } \Delta \mathrm{CQD} \end{align}

\begin{align}\angle \mathrm{APB}&=\angle \mathrm{CQD}\begin{bmatrix} \text {Each angle} \\ \text{measures } 90^{\circ} \end{bmatrix} \\ {AB}&\!=\!{CD} \begin{bmatrix}\text {Opposite sides of} \\\text{parallelogram } {ABCD}\end{bmatrix} \\ \angle \mathrm{ABP}&\!=\!\angle \mathrm{CDQ} \begin{bmatrix}\text {Alternate interior}\\ \text{angles for AB } \| \mathrm{CD}\!\end{bmatrix} \\ \therefore \Delta \mathrm{APB} &\cong \Delta \mathrm{CQD}\begin{bmatrix}\text { By AAS }\\\text{congruence rule }\end{bmatrix} \end{align}

(ii) By using the above result

\begin{align} \triangle \mathrm{APB} \cong \Delta \mathrm{CQD}, \end{align} we obtain \begin{align} \mathrm{AP}=\mathrm{CQ}\end{align} (By CPCT )

Video Solution