In rectangle $ABCD$, $AB = 2BC$. On side $AB$, point $P$ is given such that $\angle APD = \angle DPC$. Calculate the measure of this angle.

1. Problem Description

In rectangle $ABCD$, $AB = 2BC$. On side $AB$, point $P$ is given such that $\angle APD = \angle DPC$. Calculate the measure of this angle.

2. Solution Steps

Let $BC = a$, then $AB = 2a$. Let $AP = x$, then $PB = 2a - x$. Since $ABCD$ is a rectangle, $AD = BC = a$ and $CD = AB = 2a$. Let $\angle APD = \angle DPC = \theta$.

Also, let $\angle BPC = \alpha$. Since the angles around point $P$ sum to $360^\circ$, we have $\angle APD + \angle DPC + \angle BPC + \angle APB = 360^\circ$, which means $\theta + \theta + \alpha + \alpha = 360^\circ$, or $2\theta + 2\alpha = 360^\circ$. Thus, $\theta + \alpha = 180^\circ$. This implies $\angle BPC = \alpha = 180^\circ - \theta$.

Now, consider the triangles $APD$, $DPC$, and $BPC$.

In $\triangle APD$, we have $AP = x$, $AD = a$, and $PD = \sqrt{AP^2 + AD^2} = \sqrt{x^2 + a^2}$. Also, $\tan(\angle ADP) = \frac{AP}{AD} = \frac{x}{a}$.

In $\triangle BPC$, we have $PB = 2a-x$, $BC = a$, and $PC = \sqrt{PB^2 + BC^2} = \sqrt{(2a-x)^2 + a^2}$. Also, $\tan(\angle BCP) = \frac{PB}{BC} = \frac{2a-x}{a}$.

In $\triangle DPC$, we have $CD = 2a$, $DP = \sqrt{x^2 + a^2}$, $PC = \sqrt{(2a-x)^2 + a^2}$, and $\angle DPC = \theta$.

Using the Law of Cosines on $\triangle DPC$, we have

$CD^2 = DP^2 + PC^2 - 2(DP)(PC)\cos(\theta)$

$(2a)^2 = (x^2 + a^2) + ((2a-x)^2 + a^2) - 2\sqrt{(x^2+a^2)((2a-x)^2+a^2)}\cos(\theta)$

$4a^2 = x^2 + a^2 + 4a^2 - 4ax + x^2 + a^2 - 2\sqrt{(x^2+a^2)(4a^2-4ax+x^2+a^2)}\cos(\theta)$

$4a^2 = 2x^2 - 4ax + 6a^2 - 2\sqrt{(x^2+a^2)(x^2-4ax+5a^2)}\cos(\theta)$

$2\sqrt{(x^2+a^2)(x^2-4ax+5a^2)}\cos(\theta) = 2x^2 - 4ax + 2a^2$

$\sqrt{(x^2+a^2)(x^2-4ax+5a^2)}\cos(\theta) = x^2 - 2ax + a^2 = (x-a)^2$

Now, $\angle APD = \angle DPC$ implies that $PD$ and $PC$ are equally inclined to the line through $P$ perpendicular to $AB$. This implies $P$ is the midpoint of $AB$.

Let $P$ be the midpoint of $AB$, then $AP = PB = a$.

$DP = \sqrt{a^2+a^2} = a\sqrt{2}$.

$PC = \sqrt{a^2+a^2} = a\sqrt{2}$.

Thus, $DP = PC = a\sqrt{2}$. This means $\triangle DPC$ is an isosceles triangle.

Also, $CD = 2a$.

Using the Law of Cosines in $\triangle DPC$, we have

$(2a)^2 = (a\sqrt{2})^2 + (a\sqrt{2})^2 - 2(a\sqrt{2})(a\sqrt{2})\cos(\theta)$

$4a^2 = 2a^2 + 2a^2 - 4a^2\cos(\theta)$

$4a^2 = 4a^2 - 4a^2\cos(\theta)$

$0 = -4a^2\cos(\theta)$

$\cos(\theta) = 0$

$\theta = 90^\circ$.

3. Final Answer

90 degrees