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Question
Two circles are drawn with sides AB, AC of a triangle ABC as diameters. They intersect at a point D. Prove that D lies on BC.
Solution 1
AB and AC are diameters of circles with oentre O and O1 respectively
∠ ADB = 90 ° ---( 1) (Angle in a semi circle is a right angle)
Similarly, ∠ ADB = 90° ---(2)
Adding ( 1) and (2)
∠ ADB + ∠ ADC = 90 + 90
∠ BDC = 180°
Hence, BDC is a straight line.
Solution 2
Join AD.
Since angle in a semi-circle is a right angle.
Therefore,
∠ADB = 90° and ∠ADC = 90°
⇒ ∠ADB + ∠ADC = 90° + 90°
⇒ ∠ADB + ∠ADC = 180°
BDC is a straight line.
⇒ D lies on BC.
Hence proved.
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