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Figure 6, two circles touch eåch other at A. A common tangent tom at B and C and another common tangent at A meets the prenmon tangentat P. Prove that < BAC = 90°.,Figure 610 |
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Answer» Given X and Y are two circles touch each other externally at P. AB is the common tangent to the circles X and Y at point A and B respectively. To find : ∠APB Proof: let ∠CAP = α and ∠CBP = β. CA = CP [lengths of the tangents from an external point C] In a triangle PAC, ∠CAP = ∠APC = α similarly CB = CP and ∠CPB = ∠PBC = β now in the triangle APB, ∠PAB + ∠PBA + ∠APB = 180° [sum of the interior angles in a triangle] α + β + (α + β) = 180° 2α + 2β = 180° α + β = 90° ∴ ∠APB = α + β = 90° |
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