What is the area of the triangle?

Question

I managed to solve this question using trigonometry. But I wondered if there'd be anyway of doing it using only synthetic geometry. Here it is.

Let $ABC$ be a right isosceles triangle of hypotenuse $AB$. Let also $\Gamma$ be the semicircle whose diameter is the line segment $AC$ such that $\Gamma\cap\overline{AB} = \{A\}$. Consider $P\in\Gamma$ with $PC = k$, with $k \leq AC$. Find the area of triangle $PBC$.

Here is my interpretation of the picture:

I managed to get the solution via trigonometry as below.

Then, the area $S$ requested is:

$$\begin{align} S &= \displaystyle\frac{PC\cdot BC\cdot \sin(90^\circ + \beta)}{2}\\ &= \displaystyle\frac{k\cdot d\cdot \cos\beta}{2}\\ &= \displaystyle\frac{k\cdot d\cdot \frac{k}{d}}{2}\\ &= \displaystyle\frac{k^2}{2}.\\ \end{align}$$

Answer 1

$$|\triangle PBC| = \frac12|CP||BQ| = \frac12k^2$$

Answer 2

Let us find the altitude of point $P$, a is a leg. Intersection of the circles $x^2+h^2=k^2, x^2+(h-\frac{a}{2})^2=(\frac{a}{2})^2 $

$h=\frac{k^2}{a} $

$A\; PBC=\frac{1}{2}*ah=\frac{k^2}{2}$

Answer 3

Draw the parallel line BX || AP and extend PC to meet it at X. Note that $\angle XCB = 90 - \beta = \gamma$. Along with CA = CB, the right triangles APC and CBX are congruent, which leads to $CP = BX = k$.

Thus, the area of CPB is

$$Area_{CPB}=\frac12 CP\cdot BX = \frac12k^2$$

Answer 4

Let $PD$ be a perpendicular from $P$ to the line $BC$.

Thus, since a line $BC$ is a tangent to the circle and $\measuredangle DCP=\measuredangle PAC$, we see that $\Delta DCP\sim\Delta PAC,$

which gives: $$\frac{PD}{k}=\frac{k}{AC}$$ or $$PD=\frac{k^2}{AC}.$$ Id est, $$S_{\Delta PBC}=\frac{1}{2}BC\cdot PD=\frac{1}{2}BC\cdot\frac{k^2}{AC}=\frac{k^2}{2}.$$