Question

The base BC of triangle ABC Is divided at D such that BD=½DC prove that ar( triangle ABD)= ⅓×ar(triangle ABC)???​

Answer 1

Step-by-step explanation:

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

ANSWER:-

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