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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:
(ii) angles APB = 90°
Solution
ii) Now in Δ ATP ,
∴ `∠`TAP = `∠`TPA
Similarly in Δ BTP,`∠`TBP = `∠`TPB
Adding,
`∠`TAP +`∠`TBP =`∠`APB
But
∴ TAP + `∠`TBP + `∠`APB =180°
⇒ `∠`APB = `∠`TAP + `∠`TBP =90°
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