The base BC of △ ABC is divided at D such that BD=12DC..Prove that ar(△ ABD)=13ar(△ ABC).
GIVEN: BD = 1/2 DC
CONST: Draw a line ⊥ BC, AE ⊥ BC
Let AE = h (Height of ΔABC)
∵ BD + DC = BC
BD + 2BD = BC
3BD = BC
BD = 1/3 BC
Ar. ΔABD = 1/2 × BD × AE = 1/2 × (1/3 BC) × AE
= 1/3 × 1/2 x BC x AE
= 1/3 × Ar. ABC