Ex.8.1 Q10 Quadrilaterals Solution - NCERT Maths Class 9

Go back to  'Ex.8.1'

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
Quadrilaterals
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
Quadrilaterals
Ex 8.1 | Question 10