Question

In ΔPQR ,ray QD bisects ∟PQR and ray RE bisects ∟PRQ. If segPQ = seg PR, then
prove that ED II QR.

Answer 1

Answer:

In △PQR, ray QS bisects ∠PQR∴∠PQS=∠RQS

By angle bisector theorem.

$\huge \frac{pq}{ps} = \frac{qr}{rs}$

Let h be a perpendicular drawn from vertex Q to side PR

∴ Area of △PQR =

$\frac{1}{2}$

×PS×h and Area

Thus,

$\frac{= 1}{2}$

$\huge \frac{=pq}{qr}$

Be Brainly "