Consider a triangle ABC Let the angle bisector of angle alph

Consider a triangle ABC. Let the angle bisector of angle alpha interest side BC at a point D between B and C. Use the law of sines to show BD/DC = AB/AC

Solution

First we have to show that the two triangles that is ADB and ADC are similar

Here <BAD congruent to <CAD since AD is the bisector

and <ADB=180-<ADC

taking sin on both sides

sin<ADB=sin(180-<ADC)

and sin(180-theta)=sin theta

therefore sin <ADB=sin<ADC

<ADB=<ADC

therefore by AA criteria the triangle are similar

Hence BD/DC=AB/AC ( when triangles are similar their sides are in proportion)