Question

Triangle ABC is an isosceles triangle AB=Ac side BA is produced to D such that AD=AB show that angle BCD is a right angle​

Answer 1

Step-by-step explanation:

$\huge \boxed{\underbrace{\mathfrak {\underline {\fcolorbox {black}{pink}{Answer}}}}}$

${ Here....\:AB=AC }$

${ \angle{ACD}=\angle {ABC}...........(1) }$

$\implies { AB=AD }$

$\implies { so.......AC=AD }$

in ${ \triangle {ADC} }$WE have..

${ AD=AC }$

$\implies {\angle {ACD}=\angle {ADC}.......(2) }$

adding equation1 and equation2,we get.

${ \angle {ACB}+\angle{ACD}=\angle {ABC}+\angle {ADC} }$

$\implies { \angle {BAC}=\angle {ABC}+\angle {ADC} }$

Now in ${ \triangle {BCD} }$,we have..

$\boxed { \angle {BCD}+\angle {ABC}+\angle {ADC}=180° }$

$\implies { \angle {BCD}+\angle {BCD}=180° }$

$\implies {2\angle {BCD}=180° }$

$\implies { \angle {BCD}=\frac{180°}{2} }$

$\implies {\angle {BCD}=90° }$

$\boxed{\red {so\:it\:is\:a\:right-angle \:teangle} }$

Answer 2

Step-by-step explanation:

${ Here....\:AB=AC }Here....AB=AC</p><p>{ \angle{ACD}=\angle {ABC}...........(1) }∠ACD=∠ABC...........(1)</p><p>\implies { AB=AD }⟹AB=AD</p><p>\implies { so.......AC=AD }⟹so.......AC=AD</p><p>in { \triangle {ADC} }△ADC WE have..</p><p>{ AD=AC }AD=AC</p><p>\implies {\angle {ACD}=\angle {ADC}.......(2) }⟹∠ACD=∠ADC.......(2)</p><p>adding equation1 and equation2,we get.</p><p>{ \angle {ACB}+\angle{ACD}=\angle {ABC}+\angle {ADC} }∠ACB+∠ACD=∠ABC+∠ADC</p><p>\implies { \angle {BAC}=\angle {ABC}+\angle {ADC} }⟹∠BAC=∠ABC+∠ADC</p><p>Now in { \triangle {BCD} }△BCD ,we have..</p><p>\boxed { \angle {BCD}+\angle {ABC}+\angle {ADC}=180° }∠BCD+∠ABC+∠ADC=180°</p><p>\implies { \angle {BCD}+\angle {BCD}=180° }⟹∠BCD+∠BCD=180°</p><p>\implies {2\angle {BCD}=180° }⟹2∠BCD=180°</p><p>\implies { \angle {BCD}=\frac{180°}{2} }⟹∠BCD=2180°</p><p>\implies {\angle {BCD}=90° }⟹∠BCD=90°</p><p>\boxed{\red {so\:it\:is\:a\:right-angle \:teangle} }soitisaright−angleteangle</p><p>$