Question

triangle ABC is a right triangle right angled at A.. such that AB = AC.. and bisector of angle C intersect the side of AB at D... prove that AC + AD= BC

anyone else please solve this...​

Answer 1

$\sf\large{\underline{\underline{Let:}}}$

$\sf{ \: \: \: \: \: AB = AC = "x"}$

$\sf\large{\underline{\underline{ In\:Triangle\:ABC,}}}$

$\sf{\implies Sin\:( \angle B ) = \dfrac{AC}{BC}}$

$\sf{\implies Sin\:({45}\degree) = \dfrac{x}{BC}}$

$\sf{\implies \dfrac{1}{\sqrt{2}} = \dfrac{x}{BC}}$

$\sf{\implies BC = \sqrt{2}x}$

$\sf\large{CD\:is\:the\:bisector\:of\:Angle\:ACB}$

$\sf{\therefore{\dfrac{BC}{AC} = \dfrac{BD}{AD}}}$

$\sf{\implies \dfrac{\sqrt{2}x}{x} = \dfrac{AB - AD}{AD}}$

$\sf{\implies \sqrt{2} = \dfrac{x}{AD} - 1}$

$\sf{\implies \sqrt{2} + 1 = \dfrac{x}{AD}}$

$\sf{\implies AD = \dfrac{x}{\sqrt{2} + 1} × \dfrac{(\sqrt{2} -1)}{(\sqrt{2} - 1)}}$

$\sf{\implies AD = \sqrt{2}x - x }$

$\sf{\underbrace{\purple{ AD + AD = x + \sqrt{2}x - x = \sqrt{2}x = BC}}}$

$\large{\bold{\fbox{\color{red}{\: \: \: \: \: \: \: \: \: \: \: \: Hence\:Proved\: \: \: \: \: \: \: \: \: \: \: \: }}}}$