Question

In ΔABC, if AD is the bisector of ∠A, prove that Area (ΔABD)/Area (ΔACD) =AB/AC

Answer 1

Answer ;

consider a triangle ABC and an angle bisector drawn to angle A.

In Δ BAD and ΔCAD ,

∠BAD =∠CAD (given)

∠ADC=∠ADB (GIVEN)

ΔBAD≈ΔCAD (AA similarity)

by CPST , AB/AC =AB/CD

By areas of similarity triangles,

ar(ΔABD)/ar(ΔACD)=AB/AC

HENCE PROVED

Answer 2

By SSS test,

...

Let, this be figure of a question.

AD is altitude on BC.

In an isosceles triangle altitude, median and angle bisector on non equal sides are same line.

So, AD is also a median.

So, D is midpoint of BC,

⇒BD=CD

AB=AC⋯⋯[given]

And AD=AD⋯⋯[common side]

So, by SSS test

ΔABD≅ΔACD.