Ex.6.3 Q16 Triangles Solution - NCERT Maths Class 10

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Question

If AD and PM are medians of triangles ABC and PQR, respectively where \(\Delta ABC\sim\Delta PQR\), prove that \(\begin{align}\frac{AB}{PQ}=\frac{AD}{PM} \end{align} \)·

Diagram

 Video Solution
Triangles
Ex 6.3 | Question 16

Text Solution

Reasoning:

As we know if one angle of a triangle is equal to one angle of the other triangle and the sides including these angles are proportional, then the two triangles are similar.

This is referred as SAS (Side–Angle–Side) Criterion for two triangles.

Steps:

\[\begin{align}&\Delta A B C \sim \Delta P Q R\\\\\Rightarrow \quad&\angle A B C=\angle P Q R \\ &\text{(corresponding angles)} \dots(1)\\\\ \Rightarrow \quad &\;\;\;\,\frac{A B}{P Q}=\frac{B C}{Q R} \\& \text{(corresponding sides)} \\\\\Rightarrow \quad & \;\;\;\,\frac{A B}{P Q}=\frac{B C / 2}{Q R / 2} \\\\\Rightarrow \quad & \;\;\;\,\frac{AB}{PQ}=\frac{BD}{QM} \\ & \begin{bmatrix}D\text{ and }M \text{are mid points } \\ \text{of } BC \text{ and }QR\end{bmatrix} \dots(2) \end{align}\]

 

In \(\Delta ABD,\Delta PQM\)

\[\begin{align}\angle A B D=\angle P Q M\;\;\dots \text{(from 1)}\end{align}\]

\[\begin{align}\frac{A B}{P Q}=\frac{B D}{Q M}\;\;\dots \text{(from 2)}\end{align}\]

\[\begin{align}\Rightarrow \quad&\Delta A B D \sim \Delta P Q M \\& \text{(SAS Criterion)}\end{align}\]
 

\[\begin{align}\Rightarrow \quad & \frac{A B}{P Q}=\frac{B D}{Q M}=\frac{A D}{P M}\\& \text{(Corresponding sides)}\end{align}\]

\[\begin{align}  \Rightarrow \frac{A B}{P Q}&=\frac{A D}{P M}\end{align}\]