Question

1. O is a point on side PQ of a ∆PQR such that PO= QO=RO
then.

(I) RS²=PR×QR
(II)PR²+QR²=PQ²
(III)QR²=QO²+RO²
(IV)PO²+RO²=PR²​

Answer 1

${PR}^{2} + {QR}^{2} = {PQ}^{2}$

Step-by-step explanation:

PO=QO=RO (Given)

In △POR

PO=RO (Given)

∠P=∠y (1/2 R) (opposite sides and angles of an isosceles △ are equal)

In △OQR

OQ=OR

∠Q=x (1/2 R) (opposite sides and angles of an isosceles △ are equal)

x+x+y+y=180°

2x+2y=180°

2(x+y)=180°

x+y=90°

Since ∠R=90°

Using pythagoras theorem

${PR}^{2} + {QR}^{2} = {PQ}^{2}$