## In triangle PQR ,if QS is angle bisector of angle Q , then show that A( triangle PQS) upon A( triangle QRS) is equal to PQ upon QR

Question

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## Answers ( )

Step-by-step explanation:in Δ pqr if qs bisect ∠q then Area of Triangle sqr / Area of Triangle pqs = qr/pq

Step-by-step explanation:

Correct Question : A(triangle sqr)/A(triangle pqs)= qr/pq

Let say Angle q = 2β

Then qs is bisector of ∠q

=> ∠pqs = ∠rqs = β

now Draw sm ⊥ pq & sn ⊥ qr

Area of Triangle pqs

= (1/2) * pq * sm

in Δ smq

sinβ = sm/qs

=> sm = qs * sinβ

Area of Triangle pqs = (1/2) * pq *qs * sinβ

Area of Triangle sqr

= (1/2) * qr * sn

in Δ snq

sinβ = sn/qs

=> sn = qs * sinβ

Area of Triangle sqr = (1/2) * qr *qs * sinβ

Area of Triangle sqr / Area of Triangle pqs = (1/2) * qr *qs * sinβ / (1/2) * pq *qs * sinβ

=> Area of Triangle sqr / Area of Triangle pqs = qr/pq

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