In Figure 6, two circles touch eĂĄch other at A. A common tangent to

1.	In Figure 6, two circles touch eĂĄch other at A. A common tangent touchesthem at B and C and another common tangent at A meets the previouscommon tangent at P. Prove that BAC = 90Â°.Figure 630/3"10
Answer» Question is same just nomenclature is changed 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°

In Figure 6, two circles touch eĂĄch other at A. A common tangent touchesthem at B and C and another common tangent at A meets the previouscommon tangent at P. Prove that BAC = 90Â°.Figure 630/3"10

Answer»

Question is same just nomenclature is changed

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°