In following figure, center of a circle is O and tangent drawn from po

1.	In following figure, center of a circle is O and tangent drawn from point P, PA and PB which touches the circle at A and B respectively, then prove that OP line segment is bisector of AB.
Answer» Let OP intersect chord AB at point C. To prove : AC = BC. Proof : In ΔPCA and ΔPCB PA = PB (tangents at circle from external point P.) PC = PC (common side) ∠APC = ∠BPC [∵ tangents PA and PB, make equal angle with OP] by SAS congruence, ΔPCA = ΔPCB ⇒ AC = BC Hence, OP bisects line segment AB.

In following figure, center of a circle is O and tangent drawn from point P, PA and PB which touches the circle at A and B respectively, then prove that OP line segment is bisector of AB.

Answer»

Let OP intersect chord AB at point C.

To prove : AC = BC.

Proof : In ΔPCA and ΔPCB

PA = PB (tangents at circle from external point P.)

PC = PC (common side)

∠APC = ∠BPC [∵ tangents PA and PB, make equal angle with OP]

by SAS congruence,

ΔPCA = ΔPCB

⇒ AC = BC

Hence, OP bisects line segment AB.