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