Q. 75.0( 1 Vote )

# 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°

2PSR + P = 180°

P = 180° - 2PSR……(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° - 2PSR = 180°

2PSR = PQR + PRQ + 180° - 180°

2PSR = 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

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