Ex.6.3 Q16 Triangles Solution - NCERT Maths Class 10

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


 Video Solution
Ex 6.3 | Question 16

Text Solution


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.


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


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

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

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

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

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

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