Question

ABC is a right angle with AB=AC.Bisectors of angle A meets BC at angle D.Prove that BC=2AD.​

plzz help...

Answer 1

In ∆ABD and ∆ACD

AB = AC (given)

∠BAD = ∠CAD (As AD is bisector of ∠A)

AD = AD

⇒ ∆DAB = ∆DAC (by SAS congruency rule)

⇒ ∠ADB = ∠ADC (by c.p.c.t)

⇒ ∠ADB = ∠ADC = 90° and BD = DC

In ∆ABD, AD2 + BD2 = AB2 …(i)

⇒ AD2 + DC2 = AC2 …(ii)

Adding (i) and (ii), we get

2 AD2 + BD2 + DC2 = AB2 + AC2

⇒ 2AD2 + BD2 + DC2 = BC2

⇒ 2 AD2 + 2BD2 = BC2

⇒ 2 (AD2 + BD2) = BC2

⇒ 4 AD2 = BC2

⇒2 AD = BC

BC = 2AD

Hence proved.