In the given figure, Point A is a common point pf contact of two externally touching circles and line l is a common tangent to both circles touching at B and C.Line L is common tangent at A and it intersects BC at D.
Prove that(1) $∠ BAC = 90 °$ (2) Point D is the mid point of seg BC

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Solution

(i)

$DB = DA (Tangents from external point are equal) ∠ ABD = ∠ BAD . . . (i) (Angles opposite to equal sides) DC = DA (Tangents from external point are equal) ∠ DCA = ∠ CAD . . . (ii) (Angles opposite to equal sides) In △ BAC, we have : ∠ BAC + ∠ ABC + ∠ BCA = 180 ° (Angle sum property) \Rightarrow ∠ BAD + ∠ DAC + ∠ ABC + ∠ BCA = 180 ° \Rightarrow ∠ BAD + ∠ DAC + ∠ BAD + ∠ DAC = 180 ° [Using equation (i) and (ii)] \Rightarrow 2 (∠ BAD + ∠ DAC) = 180 ° \Rightarrow ∠ BAD + ∠ DAC = 90 ° \Rightarrow ∠ BAC = 90 °$

(ii)

$DB = DA . . . (i) (Tangents from external point are equal) DC = DA . . . (ii) (Tangents from external point are equal) From equations (i) and (ii), we have : DB = DC ∴ D is the midpoint of BC .$

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In the given figure, Point A is a common point pf contact of two externally touching circles and line l is a common tangent to both circles touching at B and C.Line L is common tangent at A and it intersects BC at D. Prove that(1) ∠BAC=90° (2) Point D is the mid point of seg BC

In the given figure, Point A is a common point pf contact of two externally touching circles and line l is a common tangent to both circles touching at B and C.Line L is common tangent at A and it intersects BC at D.
Prove that(1) $∠ BAC = 90 °$ (2) Point D is the mid point of seg BC