D is a point on side …
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Sia ? 6 years, 6 months ago
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} BAC
Construction: From BA produce cut off AE = A. Join CE
Proof:{tex}\frac{{BD}}{{CD}} = \frac{{AB}}{{AC}}{/tex} ........Given
{tex}\Rightarrow {/tex} {tex}\frac{{BD}}{{CD}} = \frac{{AB}}{{AE}}{/tex} {tex}\because {/tex} AC = AE(by construction)
{tex}\therefore {/tex} In {tex}\triangle {/tex} BCE,
AD {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 equal
Using (3),(1) and (2) gives {tex}\angle{/tex}BAD = {tex}\angle{/tex}CAD
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