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ABC is a triangle with angle b is equal to twice of angle c d is a point on BC such that AD bisects angle BAC and ADis equal to CD prove that angle BAC is equal to 72

Posted by Gurpreet Singh 6 years, 4 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 6 years, 4 months ago

Given, In {tex}\triangle{/tex}ABC,

{tex}\angle{/tex}B = 2{tex}\angle{/tex}C .......(1)

AD = CD ........(2)

AD bisects {tex}\angle{/tex}BAC.

So, {tex}\angle BAD = \angle DAC{/tex} = {tex}\frac {1} {2}{/tex}{tex}\angle A{/tex}.....(3)

Since AD = CD ,hence in {tex}\triangle ADC{/tex} ;
{tex}\angle{/tex}C = {tex}\angle{/tex}DAC.......(4) [angles opposite to equal sides are equal]
But from (1),

{tex}\angle{/tex}B = 2{tex}\angle{/tex}C

{tex}\Rightarrow{/tex} {tex}\angle{/tex}B = 2 {tex}\angle{/tex}DAC [ from (4) ]
{tex}\Rightarrow{/tex} {tex}\angle{/tex}B = {tex}\angle{/tex}A [ from (3) ] ..........(5)

Clearly , from (1) & (5) ;

{tex}\angle A =\angle B =2 \angle C{/tex}.......(6)

Now, {tex}\angle{/tex}A + {tex}\angle{/tex}B + {tex}\angle{/tex}C = 180° [Angle Sum Property of triangle]
{tex}\Rightarrow \angle A + \angle A + \frac{\angle A}{2} = 180°{/tex}
{tex}\Rightarrow{/tex} {tex}\frac{4\angle A + \angle A}{2} =180°{/tex}
{tex}\Rightarrow{/tex}{tex}\frac{5\angle A}{2} = 180°{/tex} {tex}\Rightarrow{/tex} {tex}\angle A{/tex} = {tex}\frac{180^o \times 2}{5}{/tex}
{tex}\Rightarrow{/tex}{tex}\angle{/tex}A = 72^o {tex}\Rightarrow{/tex} {tex}\angle{/tex}BAC = 72^o

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