triangle ABC is a right angled triangle such that Ab=BC. Bisector of angle C intersects AB at D.Prove that AC+AD=BC.

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$$\begin{gathered} Let AB = AC = a and AD = b \\ In a right angled triangle ABC , \\ B{C}^2 = A{B}^2 + A{C}^2 \\ B{C}^2 = a^2 + a^2 \\ BC = a\sqrt{2} \\ Given AD = b, we get \\ DB = AB – AD or DB = a – b \\ We have to prove that AC + AD = BC or (a + b) = a\sqrt{2}. \\ By the angle bisector theorem, we get \\ \frac{AD}{DB} =\frac{AC}{BC} \\ \frac{b}{(a - b)} =\frac{a}{a\sqrt{2}} \\ \frac{b}{\left(a-b\right)}=\frac{1}{\sqrt{2}} \\ b= \frac{(a – b)}{\sqrt{2}} \\ b\sqrt{2}=a-b \\ b\sqrt{2}+b=a \\ b\left(\sqrt{2}+1\right)=a \\ b=\frac{a}{\left(\sqrt{2}+1\right)} \\ Rationalizing the denominator with (\sqrt{2}-1) \\ b = \frac{a}{\left(\sqrt{2}+1\right)}\times \frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)} \\ b =\frac{a\left(\sqrt{2}-1\right) }{2-1} \\ b =a\left(\sqrt{2}-1\right) \\ b = a\sqrt{2} – a \\ b + a = a\sqrt{2} \\ or \mathit{A}\mathit{D}\mathbf{ }\mathbf{+}\mathbf{ }\mathit{A}\mathit{C}\mathbf{ }\mathbf{=}\mathbf{ }\mathit{B}\mathit{C}\mathbf{ } [we know that AC = a, AD = b and BC = a\surd 2] \\ Hence it is proved. \end{gathered}$$

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