What is the area of $\triangle ABC$ where $\triangle ADC$ is cyclic, point $P$ is on the circumference and $AD = AP$? 
$\triangle ABC$ is a right angled triangle. The perpendicular drawn form $A$ on $BC$ intersects $BC$ at point $D$. A point $P$ is chosen on the circle drawn through the vertices of $\triangle ADC$ such that $CP$ $\perp$ $BC$ and $AP$ = $AD$.A square is drawn on the side $BP$ and its area is 350 unit$^\text{2}$. What is the area of $\triangle ABC$?


My Attempt:
Here, $ADCP$ is a square because $AD = AP$ and $\angle ADC$ = $\angle DCP$ = 90$^\circ$. So, by expressing $AD = x$ and $BD$ = $y$, 
From the right angled $\triangle ADB$,
$AB^\text{2}$ = $x^\text{2} + y^\text{2}$
And now, from the $\triangle ABC$, 
$x^\text{2} + y^\text{2}$ + ($\sqrt 2 x$)$^\text{2}$ = $(x + y)$$^\text{2}$.....($AC$ is the diagonal of the square $ADCP$)
$x^\text{2}$ + $y^\text{2}$ + 2x$^\text{2}$ = $x^\text{2}$ + $y^\text{2}$ + 2$xy$ $\implies$ $x$ =$y$
So, $BD$ = $x$ and $AB$ = $\sqrt(2x^\text{2})$ $\implies$ $AB$ = $\sqrt2 x$. And now from $\triangle BCP$,
($2x$)$^\text{2}$ + $x$$^\text{2}$ = ($\sqrt350$)$^\text{2}$
$5x$$^\text{2}$ = $350$ $\implies$ $x$$^\text{2}$ = $70$ $\implies$ $x$ = $\sqrt70$
After that, the area of $\triangle ABC$ = $\frac{1}{2}$×$\sqrt2x$×$\sqrt2x$ 
= $\frac {1}{2}$×$2x$ $^\text{2}$ = $x^\text{2}$ = ($\sqrt70$)$^\text{2}$ = $70$.
Is my answer correct? If not so, can anyone please provide me with another solution or method for better learning process? Or simply any kind of clue or hint will be so much helpful. Thanks in advance.

Since $AC$ is the diagonal of square $APCD$, then $\angle{C}=45^{\circ}$. Then, we know $\triangle{ABC}$ is a 45-45-90 right triangle. Notice that $ABC$ is isosceles, so $BD=DC=CB=x$. Then, you can use Pythagorean Theorem to find $x$. Then, it is easily seen that $[ABC]=\frac{(\sqrt{2\cdot 70})^2}{2}=70$. 
