Question

in PQR QPR=90 PA perpendicilular QR
AB PERPENDICULAR PR
TO PROVE PB×PR=QA×AR​

Answer 1

Answer:

T×(PQ+PR)=PQ×PR.

Step-by-step explanation:

Given :

Angle QPR=90°,

PS is it's bisector

=> ∠QPS = ∠RPS = 90°/2 = 45°

ST ⊥ PR

=> in Δ PST

∠STP = 90°  & ∠TPS = ∠RPS = 45°  ( as T Lies on PR)

=> ∠TSP = 180° - ∠STP - ∠TPS = 180° - 90° - 45° = 45°

=> in Δ PST

∠TPS = ∠TSP = 45°

=> ST = PT

now in Δ STR & ΔQPR

∠STR = ∠QOR = 90°

∠R is common

so Third angle would also be equal

Hence  Δ STR ≅ ΔQPR

=> ST/PQ  = TR /PR

=> ST * PR  = PQ * TR

TR = PR - PT

while PT = ST

=> TR = PR - ST

using this value

=> ST * PR  = PQ * (PR - ST)

=> ST * PR  = PQ * PR - PQ  * ST

=> ST * PR + PQ  * ST = PQ * PR

=> ST(PR + PQ)  = PQ * PR

=> ST×(PQ+PR)=PQ×PR.

QED

Proved