Question

t w o chords PQ and PR of a circle are equal then prove that the bisector of angle rpq passes through the centres of circle​

Answer 1

Answer:

Given, chords RP=RQ

In △PSQ and △PSR

PQ=PR (given)

∠RPS=∠QPS (given)

PS=PS (common)

△PSQ≅△PSR (by SAS)

⇒RS=QS

∠PSR=∠PSQ

But,

∠PSR+∠PSQ=180 °

2∠PSR=180 °

∠PSQ=∠PSR=90 °

then, RS=QS and ∠PSR=90 °

PS is the perpendicular bisector of chord RQ

PS passes through center of circle.