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Q12:

Sides $AB$ and $BC$ and median $AD$ of a triangle $ABC$ are respectively proportional to sides $PQ$ and $QR$ and median $PM$ of $\triangle PQR$ (see Fig. 6.41). Show that $\triangle ABC \sim \triangle PQR$.

Solution :

Given:

In $\triangle ABC$ and $\triangle PQR$, $AD$ is the median to side $BC$ and $PM$ is the median to side $QR$.

The sides and medians are proportional such that: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$.

To Prove:

$\triangle ABC \sim \triangle PQR$

A B C D P Q R M

Step 1: Analyzing the given ratios

We are given: $\frac{AB}{PQ} = \frac{BC}{QR} = \frac{AD}{PM}$

Since $AD$ and $PM$ are medians, $D$ is the midpoint of $BC$ and $M$ is the midpoint of $QR$.

Therefore, $BC = 2BD$ and $QR = 2QM$.

Substituting these into the ratio: $\frac{BC}{QR} = \frac{2BD}{2QM} = \frac{BD}{QM}$.

Thus, the given condition becomes: $\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$.

Step 2: Proving $\triangle ABD \sim \triangle PQM$

In $\triangle ABD$ and $\triangle PQM$:

$\frac{AB}{PQ} = \frac{BD}{QM} = \frac{AD}{PM}$ [Derived in Step 1]

By the SSS (Side-Side-Side) similarity criterion, $\triangle ABD \sim \triangle PQM$.

Consequently, the corresponding angles are equal: $\angle B = \angle Q$ [Since corresponding parts of similar triangles are equal (CPST)].

Step 3: Proving $\triangle ABC \sim \triangle PQR$

In $\triangle ABC$ and $\triangle PQR$:

1. $\frac{AB}{PQ} = \frac{BC}{QR}$ [Given]

2. $\angle B = \angle Q$ [Proved in Step 2]

By the SAS (Side-Angle-Side) similarity criterion, $\triangle ABC \sim \triangle PQR$.

Final Answer:

Since the ratio of two sides is proportional and the included angle is equal, $\triangle ABC \sim \triangle PQR$ is proved.


More Questions from Class 10 Mathematics Triangles EXERCISE 6.3


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