In the given figure, O and O' are centers of two circles, touching each other at point P. The common tangent at P meets a direct common tangent AB at M. Show that :

(i) M is mid-point of AB.

(ii) ∠APB = 90°

(i) From figure,

From M, MA and MP are the tangents.

∴ MA = MP …….(1) (∵ length of the different tangents to a circle from a single point are equal.)

Similarly,

From M, MB and MP are the tangents.

∴ MB = MP ………(2) (∵ length of the different tangents to a circle from a single point are equal.)

From (1) and (2),

MA = MB.

Hence, proved that M is the mid-point of AB.

(ii) Join AP and BP.

∵ MA = MP

∴ In Δ APM,

∠MAP = ∠MPA ……….(3) (∵ angles opposite to equal sides are equal.)

∵ MB = MP

∴ In Δ BPM,

∠MPB = ∠MBP ………(4) (∵ angles opposite to equal sides are equal.)

Adding equations (3) and (4)

⇒ ∠MAP + ∠MPB = ∠MPA + ∠MBP

⇒ ∠MAP + ∠MBP = ∠APB

Since sum of angles in a triangle = 180°

In Δ APB,

⇒ ∠APB + ∠BAP + ∠ABP = 180°

From figure,

∠BAP = ∠MAP

∠ABP = ∠MBP

Putting value of ∠MAP + ∠MBP = ∠APB in above equation

⇒ ∠APB + ∠APB = 180°

⇒ 2∠APB = 180°

⇒ ∠APB = $\dfrac{180°}{2}$ = 90°.

Hence, proved that ∠APB = 90°.