Question

In given figure $\mathrm{\angle }1=\mathrm{\angle }2$ and $\mathrm{\Delta }NSQ\cong \mathrm{\Delta }MTR$ , then prove that $\mathrm{\Delta }PTS\sim \mathrm{\Delta }PRQ$

Answer 1

Hint:
We can equate the corresponding sides of congruent triangles. Then we can equate the sides opposite to equal angles in a triangle. From these two relations, we can show that 2 sides are proportional. Then we can take the common angle of the triangles. Using these conditions, we can prove the required triangles are similar.

Complete step by step solution:
Consider the triangle PST,
It is given that $\mathrm{\angle }1=\mathrm{\angle }2$ .
We know that in a triangle, the sides opposite to equal angles will be equal.
So, from the figure, we can write,
 $PT=PS$ … (1)
It is given that $\mathrm{\Delta }NSQ\cong \mathrm{\Delta }MTR$ .
We know that corresponding sides of congruent triangles are equal.
 $\Rightarrow SQ=TR$ .. (2).
Now we can add equations (2) and (1).
 $\Rightarrow PS+SQ=PT+TR$
From the figure, we can write,
 $\Rightarrow PQ=PR$ .. (3)
Now we can consider the triangles PTS and PRQ.
We can consider the angles $$\mathrm{\angle }SPT$$ and $$\mathrm{\angle }QPR$$ .
From the figure, they represent the same angles. So, they will be equal.
 $$\Rightarrow \mathrm{\angle }SPT=\mathrm{\angle }QPR$$
Now consider the sides PS and PQ. We can take their ratio.
 $$\Rightarrow {\displaystyle \frac{PS}{PQ}}$$
Now we can apply equations (1) and (3). Then we get,
  $$\Rightarrow {\displaystyle \frac{PS}{PQ}}={\displaystyle \frac{PT}{PR}}$$
So, the 2 sides are proportional.
As the 2 sides are proportional and the corresponding angle between them are equal, we can say that the triangles PTS and PRS are similar.
So, we get, $\mathrm{\Delta }PTS\sim \mathrm{\Delta }PRQ$
Hence proved.

Note:
Note: Alternate solution is given by,
 Consider the triangle PST,
It is given that $\mathrm{\angle }1=\mathrm{\angle }2$ .
We know that in a triangle, the sides opposite to equal angles will be equal.
So, from the figure, we can write,
 $PT=PS$ … (a)
It is given that $\mathrm{\Delta }NSQ\cong \mathrm{\Delta }MTR$ .
We know that corresponding sides of congruent triangles are equal.
 $\Rightarrow SQ=TR$ .. (b).
Now we can add equations (a) and (b).
 $\Rightarrow PS+SQ=PT+TR$
From the figure, we can write,
 $\Rightarrow PQ=PR$ .. (c)
PQ and PR are sides of the triangle PQR and are equal. Then the opposite angles will be equal.
 $\Rightarrow \mathrm{\angle }PQR=\mathrm{\angle }PRQ$ … (d)
Consider the triangles PTS and PRQ
By taking the angle sum property we get,
 $\Rightarrow {180}^{\circ }=\mathrm{\angle }P+\mathrm{\angle }1+\mathrm{\angle }2$
On rearranging and applying the given condition, we get,
 $\Rightarrow {180}^{\circ }-\mathrm{\angle }P=2\mathrm{\angle }1$ … (e)
 $\Rightarrow {180}^{\circ }=\mathrm{\angle }P+\mathrm{\angle }PQR+\mathrm{\angle }PRQ$
On rearranging and substituting (d), we get
 $\Rightarrow {180}^{\circ }-\mathrm{\angle }P=2\mathrm{\angle }PQR$ … (f)
On comparing (e) and (f), we get,
 $\Rightarrow \mathrm{\angle }1=\mathrm{\angle }PQR$
On substituting, equation (d) and given condition, we get,
 $\Rightarrow \mathrm{\angle }2=\mathrm{\angle }PRQ$
Now we can consider the triangles PTS and PRQ.
We can consider the angles $$\mathrm{\angle }SPT$$ and $$\mathrm{\angle }QPR$$ .
From the figure, they represent the same angles. So, they will be equal.
 $$\Rightarrow \mathrm{\angle }SPT=\mathrm{\angle }QPR$$
From the above results, we can write,
 $\Rightarrow \mathrm{\angle }1=\mathrm{\angle }PQR$
 $\Rightarrow \mathrm{\angle }2=\mathrm{\angle }PRQ$
As the corresponding angles are equal, we can say that the triangles PTS and PRS are similar.
So, we get, $\mathrm{\Delta }PTS\sim \mathrm{\Delta }PRQ$
Hence proved.