Hello, jenjen!

Given an isosceles triangle $\displaystyle ABC$ with vertices: .$\displaystyle A (0,h),\;B(-x,0),\;(x, 0)$

Then point $\displaystyle J$ where the angle bisectors meet lies on the y-axis.

Find its y-coordinate. Code:

A|
*(0,h)
/|\
/ | \
/ | \
/ | \
/ | J \
/ * \
/ θ * |y \
B / * θ | \ C
- - * - - - -+- - - - * - -
(-x,0) D (x,0)

Let $\displaystyle \theta = \angle JBC = \angle ABJ$

We know that: .$\displaystyle \tan2\theta \:=\:\frac{2\tan\theta}{1-\tan^2\!\theta} \:=\:\frac{h}{x}$

. . and we see that: .$\displaystyle \tan\theta \:=\:\frac{y}{x}$

So we have: .$\displaystyle \frac{2\tan\theta}{1-\tan^2\!\theta} \:=\:\frac{h}{x}\quad\Rightarrow\quad \frac{2\left(\frac{y}{x}\right)}{1-\left(\frac{y}{x}\right)^2} \:=\:\frac{h}{x}$

This simplifies to a quadratic in $\displaystyle y\!:\;\;hy^2 + 2x^2y - hx^2\:=\:0$

Quadratic Formula: .$\displaystyle y \;=\;\frac{-2x^2\pm\sqrt{4x^4 + 4h^2x^2}}{2h} \;=\;\frac{-x^2\pm x\sqrt{x^2+h^2}}{h} $

. . and the positive root is: .$\displaystyle y \;=\;\frac{x\sqrt{x^2+h^2} - x^2}{h}$