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

(i) tangent at point P bisects AB.

(ii) angle APB = 90°.

(i) We know that,

If two tangents are drawn to a circle from an exterior point, the tangents are equal in length.

From figure,

TA and TP are the tangents to the circle with center O.

∴ TA = TP ………..(1)

TB and TP are the tangents to the circle with center O'.

∴ TB = TP ………..(2)

From (1) and (2) we get :

TA = TB.

Hence, proved that tangent at point P bisects AB.

(ii) In △ATP,

TA = TP [Proved above]

∴ ∠TAP = ∠TPA ………(1) [∵ angles opposite to equal sides are equal.]

In △BTP,

TB = TP [Proved above]

∴ ∠TBP = ∠TPB ……….(2) [∵ angles opposite to equal sides are equal.]

Adding (1) and (2), we get :

∠TAP + ∠TBP = ∠TPA + ∠TPB

∠TAP + ∠TBP = ∠APB ………..(3)

In △ABP,

⇒ ∠APB + ∠BAP + ∠ABP = 180° [Angle sum property of triangle]

⇒ ∠APB + ∠TAP + ∠TBP = 180° [From figure, ∠TAP = ∠BAP and ∠TBP = ∠ABP.]

⇒ ∠APB + ∠APB = 180°

⇒ 2∠APB = 180°

⇒ ∠APB = 90°.

Hence, proved that ∠APB = 90°.