Question

We have to prove the theorem that if in two triangles, corresponding angles are equal then their corresponding sides are in the same ratio (or proportion) and hence the two triangles are similar.

Answer 1

Hint: First of all draw two triangles ABC and PQR and then mark two points D and E on the side PQ and PR such that $PD=AB$ and $PR=AC$. Then prove the congruence of two triangles ABC and PDE then by CPCT we can show that angle A is equal to angle D and it is already given that angle A and angle Q are equal so from this we can deduce that angle D is equal to angle Q. Similarly, we can show angle E is equal to angle R. Hence, we have shown that DE is parallel to QR. Now, using the theorem that, if a line (DE) is parallel to one side of the triangle and intersect the other two sides in two distinct points D and E then the other two sides are divided in the same ratio, we can shown that the sides of the triangle ABC and PQR are proportional to each other. Now, if sides of the two triangles are proportional then the two triangles are similar.

Complete step by step answer:
In the below figure, we have drawn two triangles ABC and PQR and we have drawn DE on PQR as follows:

In the above figure, $AB=PD\And AC=PE$.
It is given that corresponding angles of triangles ABC and PQR are equal so:
$\angle A=\angle P,\angle B=\angle Q,\angle C=\angle R$
The below sides are equal due to its construction in such a way:
 $AB=PD\And AC=PE$
Let us consider $\Delta ABC\And \Delta PDE$,
$AB=PD\And AC=PE$
$\angle A=\angle P$
Then $\Delta ABC\cong \Delta PDE$ by SAS congruence rule.
As $\Delta ABC\cong \Delta PDE$ then the following angles are equal due to CPCT(Corresponding part of Congruent triangles),
$\begin{align}
  & \angle B=\angle D........Eq.(1) \\
 & \angle C=\angle E........Eq.(2) \\
\end{align}$
It is also given that:
$\begin{align}
  & \angle B=\angle Q...........Eq.(3) \\
 & \angle C=\angle R...........Eq.(4) \\
\end{align}$
From eq. (1 and 3) we get,
$\angle D=\angle Q$ …….. Eq. (5)
From eq. (2 and 4) we get,
$\angle E=\angle R$………. Eq. (6)
From eq. (5 and 6), we can say that DE is parallel to QR then the above angles are corresponding angles.

There is a theorem that if a line is parallel to one side of the triangle and intersects the other two sides in two distinct points then the other two sides are divided in the same ratio. Using this theorem in the triangle PQR, $DE\parallel QR$ and the side DE intersects PQ and PR in two distinct points D and E respectively.
Then, the following sides are proportional in the following way:
$\dfrac{PD}{DQ}=\dfrac{PE}{ER}$
Taking reciprocal on both the sides of the above equation we get,
$\dfrac{DQ}{PD}=\dfrac{ER}{PE}$
Adding 1 on both the sides we get,
$\begin{align}
  & \dfrac{DQ}{PD}+1=\dfrac{ER}{PE}+1 \\
 & \Rightarrow \dfrac{DQ+PD}{PD}=\dfrac{ER+PE}{PE} \\
 & \Rightarrow \dfrac{PQ}{PD}=\dfrac{PR}{PE} \\
\end{align}$
Taking reciprocal on both the sides we get,
$\dfrac{PD}{PQ}=\dfrac{PE}{PR}$ ………. Eq. (7)
We have shown above that:
$AB=PD\And AC=PE$
So, substituting $AB=PD\And AC=PE$ in eq. (7) we get,
$\dfrac{AB}{PQ}=\dfrac{AC}{PR}$
Similarly, we can show that:
$\dfrac{AB}{PQ}=\dfrac{BC}{QR}$
Therefore, assemble all the proportional sides that we have shown above we get,
$\dfrac{AB}{PQ}=\dfrac{AC}{PR}=\dfrac{BC}{QR}$
Since, we have shown that all the sides of the triangles ABC and PQR are proportional so:
$\Delta ABC\sim \Delta PQR$

Hence, we have shown that two triangles ABC and PQR are similar.

Note: From this solution, we have extracted some information that we can directly use in solving the other problems are as follows:
First of all the theorem itself states that if corresponding sides of two triangles are equal then the sides of the triangle are proportional to each other and hence, the two triangles are similar to each other.
The other thing is that if two triangles are congruent then corresponding part of the congruent triangles are equal.