In ΔPQR, PQ > PR; Lets cut off the line segment PS equal to the length of PR from the side PQ. Let’s join two points R and S. Let’s prove that

(i) ∠ PSR = 1/2 (∠ PQR + ∠ PRQ)

(ii) ∠ QRS = 1/2 (∠ PRQ - ∠ PQR)

Given: ΔPQR, PQ > PR, PS = PR

To prove: (i)

(ii)

The figure to the given question is as shown below,

(i) Now Consider the ΔPSR,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠PSR + ∠PRS + ∠P = 180°……(i)

But given PS = PR

And we know angles opposite to equal sides are equal, so

∠PSR = ∠PRS………(ii)

Substituting this value in equation (i), we get

∠PSR + ∠PSR + ∠P = 180°

⇒ 2∠PSR + ∠P = 180°

⇒ ∠P = 180° - 2∠PSR……(iii)

Now Consider the ΔPQR,

We know in a triangle the sum of all three interior angles is equal to 180°.

So in this case,

∠PQR + ∠PRQ + ∠P = 180°

Substituting the value of ∠P from equation (iii), we get

∠PQR + ∠PRQ + 180° - 2∠PSR = 180°

⇒ 2∠PSR = ∠PQR + ∠PRQ + 180° - 180°

⇒ 2∠PSR = ∠PQR + ∠PRQ

(iv)

Hence proved

(ii) From equation (ii), ∠PSR = ∠PRS

And from equation (iv),

Comparing these two we get

From figure, ∠PRS = ∠PRQ - ∠QRS

Substituting this value in equation (v), we get

Hence Proved