D is a point on side BC of triangleABC such that BD/CD=AB/AC .prove that AD is the bisector of

D is a point on side BC of triangleABC such that BD/CD=AB/AC .prove th

1.	D is a point on side BC of triangleABC such that BD/CD=AB/AC .prove that AD is the bisector of
Answer» In the figure, D is a point on side BC of {tex}\\triangle {/tex} ABC such that.{tex}\\frac{{BD}}{{CD}} = \\frac{{AB}}{{AC}}{/tex}To prove, AD is the bisector of{tex}\\angle{/tex} BACConstruction: From BA produce cut off AE = A. Join CEProof:{tex}\\frac{{BD}}{{CD}} = \\frac{{AB}}{{AC}}{/tex} ........Given{tex}\\Rightarrow {/tex}{tex}\\frac{{BD}}{{CD}} = \\frac{{AB}}{{AE}}{/tex}\xa0{tex}\\because {/tex} AC = AE(by construction){tex}\\therefore {/tex} In \u200b{tex}\\triangle {/tex}\u200b BCE,AD\xa0{tex}\\parallel{/tex} CE ............By converse of the basic proportional theorem{tex}\\therefore {/tex}{tex}\\angle{/tex}BAD = {tex}\\angle{/tex}AEC .......(1)...........Corres. {tex}\\angle{/tex} s{tex}\\angle{/tex}CAD = {tex}\\angle{/tex}AEC ..........(2) ........Alt,Int. {tex}\\angle{/tex} s{tex}\\therefore {/tex} AC = AE .........By construction{tex}\\therefore {/tex}{tex}\\angle{/tex}AEC = {tex}\\angle{/tex}ACE ............(3)...angles opposite equal sides of a triangle are equalUsing (3),(1) and (2) gives {tex}\\angle{/tex}BAD = {tex}\\angle{/tex}CAD