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In the figure (ii) given below, O and O' are centres of two circles touching each other externally at the point P. The common tangent at P meets a direct common tangent AB at M. Prove that,

(i) M bisects AB.

(ii) ∠APB = 90°.

In the figure (ii) given below, O and O' are centres of two circles touching each other externally at the point P. The common tangent at P meets a direct common tangent AB at M. Prove that (i) M bisects AB. (ii) ∠APB = 90°. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Circles

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Answer

(i) From figure,

From M, MA and MP are the tangents.

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

From (i) and (ii),

MA = MB.

Hence, proved that M bisects AB.

(ii) Since MA = MP

Hence in triangle APM,

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

Since MB = MP

Hence in triangle BPM,

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

Adding equations (i) and (ii)

⇒ ∠MAP + ∠MPB = ∠MPA + ∠MBP
⇒ ∠MAP + ∠MBP = ∠APB

Since sum of angles in a triangle = 180°

In triangle APB

⇒ ∠APB + ∠MAP + ∠MBP = 180°

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

⇒ ∠APB + ∠APB = 180°

⇒ 2∠APB = 180°

⇒ ∠APB = 180°2\dfrac{180°}{2} = 90°.

Hence, proved that ∠APB = 90°.

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