#### 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°

Is there an error in this question or solution?

Solution 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° Concept: Tangent Properties - If Two Circles Touch, the Point of Contact Lies on the Straight Line Joining Their Centers.