Pls solve this

The equal chords AB and CD of a circle with centre O,when produced,meet at P outside the circle.Prove that 1.PB=PD, 2.PA=PC.

 

Angles subtended by equal chords are equal.

So we have AC||BD, alternate angles test.

in triangle APC

BD ||AC

so

By BPT

AB/PB=CD/PD

now

AB=CD

so

PB=PD..............Ans1

Add AB on both sides

PB+AB=PD+AB

PB+AB=PD+CD........AB=CD

PA=PC...........Ans2