Question

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

Answer 1

Given : AD ⊥ BC

2DB = 3CD

Step-by-step explanation:

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.

Answer 2

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.