In rectangle $ABCD$, $AB = 2BC$. Point $P$ is on side $AB$ such that $\angle APD = \angle DPC$. Find the measure of $\angle APD$.

1. Problem Description

In rectangle $ABCD$, $AB = 2BC$. Point $P$ is on side $AB$ such that $\angle APD = \angle DPC$. Find the measure of $\angle APD$.

2. Solution Steps

Let $BC = a$, so $AB = CD = 2a$. Let $AP = x$, so $PB = 2a - x$.

Since $\angle APD = \angle DPC$, let $\angle APD = \angle DPC = \theta$. Let $\angle BPC = \alpha$. Then $\theta + \theta + \alpha = 180^{\circ}$, so $\alpha = 180^{\circ} - 2\theta$.

In $\triangle APD$, $\tan(\angle APD) = \tan \theta = \frac{AD}{AP} = \frac{a}{x}$.

In $\triangle BPC$, $\tan(\angle BPC) = \tan \alpha = \tan(180^{\circ} - 2\theta) = -\tan(2\theta) = \frac{BC}{PB} = \frac{a}{2a - x}$.

$\tan(2\theta) = \frac{2\tan \theta}{1 - \tan^2 \theta}$. Thus $-\tan(2\theta) = \frac{-2\tan \theta}{1 - \tan^2 \theta} = \frac{2\tan \theta}{\tan^2 \theta - 1}$.

Substituting $\tan \theta = \frac{a}{x}$, we get $\frac{2\frac{a}{x}}{\frac{a^2}{x^2} - 1} = \frac{2\frac{a}{x}}{\frac{a^2 - x^2}{x^2}} = \frac{2ax}{a^2 - x^2}$.

Then $\frac{a}{2a - x} = \frac{2ax}{a^2 - x^2}$, so $a(a^2 - x^2) = 2ax(2a - x)$.

Dividing by $a$, we get $a^2 - x^2 = 2x(2a - x) = 4ax - 2x^2$.

So $x^2 - 4ax + a^2 = 0$. Using the quadratic formula,

$x = \frac{4a \pm \sqrt{16a^2 - 4a^2}}{2} = \frac{4a \pm \sqrt{12a^2}}{2} = \frac{4a \pm 2a\sqrt{3}}{2} = (2 \pm \sqrt{3})a$.

Since $x < 2a$, we take the minus sign, so $x = (2 - \sqrt{3})a$.

Then $\tan \theta = \frac{a}{x} = \frac{a}{(2 - \sqrt{3})a} = \frac{1}{2 - \sqrt{3}} = \frac{2 + \sqrt{3}}{(2 - \sqrt{3})(2 + \sqrt{3})} = \frac{2 + \sqrt{3}}{4 - 3} = 2 + \sqrt{3}$.

$\theta = \arctan(2 + \sqrt{3}) = 75^{\circ}$.

3. Final Answer

The angle $\angle APD = 75^{\circ}$.