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°

 

Answer 1

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°