Question

ΔABC is a right triangle right angled at A such that AB = AC and bisector of ∠C intersects the side AB at D. Prove that AC + AD = BC.

Answer 1

Given: In right triangle ΔABC, ∠BAC = 90$°$, AB = AC and ∠ACD = ∠BCD.

To prove: AC + AD = BC

Proof:
Let AB = AC = x and AD = y.

In $∆$ABC,

$$\begin{gathered} B{C}^2=A{B}^2+A{C}^2 \\ \Rightarrow B{C}^2=x^2+x^2 \\ \Rightarrow B{C}^2=2x^2 \\ \Rightarrow BC=\sqrt{2x^2} \\ \Rightarrow BC=x\sqrt{2} \end{gathered}$$

Now,

$\frac{BD}{AD}=\frac{BC}{AC}$ (An angle bisector of an angle of a triangle divides the opposite side in two segments that are proportional to the other two sides of the triangle.)
$$\begin{gathered} \Rightarrow \frac{x-y}{y}=\frac{x\sqrt{2}}{x} \\ \Rightarrow \frac{x}{y}-\frac{y}{y}=\sqrt{2} \\ \Rightarrow \frac{x}{y}-1=\sqrt{2} \\ \Rightarrow \frac{x}{y}=\left(\sqrt{2}+1\right) \\ \Rightarrow y=\frac{x}{\left(\sqrt{2}+1\right)} \end{gathered}$$
$$\begin{gathered} \Rightarrow y=\frac{x}{\left(\sqrt{2}+1\right)}\times \frac{\left(\sqrt{2}-1\right)}{\left(\sqrt{2}-1\right)} \\ \Rightarrow y=\frac{x\left(\sqrt{2}-1\right)}{{\left(\sqrt{2}\right)}^2-1^2} \\ \Rightarrow y=\frac{x\sqrt{2}-x}{2-1} \\ \Rightarrow y=x\sqrt{2}-x \\ \Rightarrow x+y=x\sqrt{2} \end{gathered}$$

Hence, AC + AD = BC.