Question

∆ABC is an isosceles triangle in which AB = AC. Also, D is a point such that BD = CD. Prove that AD bisects angle A and angle D.

Answer 1

Given : ABC is an isosceles triangle with AB = AC and AD is the median.





In ΔABD and ΔACD

AB = AC (Given)

AD = AD (Common)

BD = DC (from (1))

ΔABD ΔACD (by SSS congurence criterion)

⇒ ∠ADB = ∠ADC

but ∠ADB = ∠ADC = 180°

⇒ ∠ADB = ∠ADC = 90°



Now in right ΔADB

AB2 = AD2 + BD2

⇒ AB2 – AD2 = BD2

⇒ AB2 – AD2 = BD × BC

⇒ AB2 – AD2 = BD × CD (from (1))


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Answer 2

AB=AC (GIVEN)
BD=CD (GIVEN)
AD=AD (COMMON)
BY SSS CONGRUENCE CRITERION
ΔADB≌ΔADC
∠BAD=∠CAD (CPCT)
∠BDA=∠CDA (CPCT)