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.

Answer 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. 

Answer 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.