Question

the base BC of triangle ABC is divided at D such that BD= 1/2DC. prove that ar.(triangle ABD) = 1/3× ar(triangle ABC)

Answer 1

GIVEN: BD = 1/2 DC
CONST: Draw a line ⊥ BC, AE ⊥ BC
Let AE = h (Ht. of ΔABC)
∵ BD + DC = BC
BD + 2BD = BC
3BD = BC
BD = 1/3 BC
Ar. ΔABD = 12 × BD × AE = 1/2 × (1/3 BC) × AE
= 13 × 12 = BC × AE
= 13 × Ar. ABC

Hope I helped you! :)

Answer 2

here is your answer I hope this helps you