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\(\frac{AB}{PQ}=\frac{AD}{PM}\)·

Diagram

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:

\(\Delta A B C \sim \Delta P Q R\)

\(\begin{align}\Rightarrow \angle A B C=\angle P Q R  \end{align}\qquad\) (corresponding angles) .........(1)

  \(\begin{align}\Rightarrow \frac{A B}{P Q}=\frac{B C}{Q R}\end{align}\qquad\)  (corresponding sides)

\(\begin{align}& {\Rightarrow \frac{A B}{P Q}=\frac{B C / 2}{Q R / 2}}\end{align}\)

\(\begin{align}\Rightarrow \frac{AB}{PQ}=\frac{BD}{QM}\end{align}\qquad\)  (D and M are mid points of BC and QR) .......(2)

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

\(\angle A B D=\angle P Q M\) (from 1)
   
\(\begin{align}\frac{A B}{P Q}=\frac{B D}{Q M}\end{align}\qquad\) (from 2) 

\(\begin{align}\Rightarrow \quad\Delta A B D \sim \Delta P Q M\end{align}\qquad\)  (\(SAS\) criterion)

\(\begin{align}\Rightarrow\quad \frac{A B}{P Q}&=\frac{B D}{Q M}=\frac{A D}{P M}\end{align}\qquad\) (corresponding sides)

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