Mathematics
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
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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 = = 90°.
Hence, proved that ∠APB = 90°.
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