in triangle ABC,D is a point in side BC such that 2BD = 3DC prove that the area of triangle ABD = 3/5 * Area of ABC
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Given : AD ⊥ BC
2DB = 3CD
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ANSWER
AD=AC ( Given )
So, ∠ACD=∠ADC [ Angles opposite to equal sides are equal ]
Now ∠ACD is exterior angle of △ABD
∠ADC=∠ABD+∠BAD [ exterior angles sum of interior opposite angles ]
So, ∠ADC>∠ABD
∠ACD>∠ABD ( from (1))
AB>AC [ side opposite to greater angle is longer ]
∴AB>AD ( As AC=AD given )
Hence, proved.
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