Question

Prove that the circle drawn on any one of the equal sides of an isosceles triangle as diameter bisects the base. 

Answer 1

To prove: BD =DC 

Proof: let .AB be the diameter of the circle with centre O 

∴ ∠ ADB = 900    (Angles in a semicirde is a right triangle) 

∠ ADB + ∠ ADC = 180     (linear pair) 

∴ ∠ ADC = 180 - 90 = 90°

In  Δ ADB and Δ ADC 

AB= AC    (given) 

∠ ADB = ∠ ADC  (90"each) 

AD= AD (Common) 

∴  Δ ADB ≅  Δ ADC   (RHS) 

Hence BD = DC    (CPCT) 

Answer 2

Given: In Isosceles Δ ABC. A circle is drawn taken AB as diameter which intersects BC at D.

To prove : BD = DC

Construction: Join AD.
Proof: ∠ADB = 90°     ...(Angle of semi-circle)
In Δ ABD and Δ ACD,
AB = AC                       ...(Given)
∠ADB = ∠ADC            ...(90°)
AD = AD                      ....(Common)
∴ Δ ABD ≅ Δ ACD
Hence, BD = DC.