Question

In the given figure, two circles touch each other externally at point P. AB is the direct common tangent of these circles. Prove that :

1. tangent at point P bisects AB,
2. angles APB = 90°.

Answer 1

Draw TPT' as common tangent to the circles.

i. TA and TP are the tangents to the circle with centre O.

Therefore, TA = TP   ...(i)

Similarly, TP = TB  ...(ii)

From (i) and (ii)

TA = TB

Therefore, TPT' is the bisector of AB.

ii. Now in ΔATP,

∴ ∠TAP = ∠TPA

Similarly in ΔBTP, ∠TBP = ∠TPB

Adding,

∠TAP +∠TBP = ∠APB

But

∴ TAP + ∠TBP + ∠APB = 180°

`=>` ∠APB = ∠TAP + ∠TBP = 90°