# 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.

• Another way is to observe $D$ is the midpoint of $BC$ and $\triangle{ABC}$ is isosceles with equal sides as $\sqrt{2\cdot 70}$ – mnulb Feb 10 '19 at 3:57

Since $$ADCP$$ is square, we obtain $$AD=BD=DC=x$$
because $$\measuredangle ACD=45^{\circ}$$ and from here also $$\measuredangle ABD=45^{\circ}.$$
Thus, $$BC=2x$$ and $$350=PB^2=(2x)^2+x^2,$$ which gives $$x^2=70$$ and $$S_{\Delta ABC}=\frac{2x\cdot x}{2}=70.$$
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 $$[ABC]=\frac{(\sqrt{2\cdot 70})^2}{2}=70$$.
• After denoting the $\triangle ABC$ as an isosceles triangle, I still get the area of $\triangle ABC$ is 70. Would you please narrate your last sentence? – Anirban Niloy Feb 10 '19 at 5:11
• Try it on geogebra. I got the area on geogebra to be $349.1$. To clarify, I multiplied by $\sqrt{2}$ because the diagonal $AC=\sqrt{2} \cdot \sqrt{350}$, due to 45-45-90 triangles. Then, I used $\frac{1}{2}bh$. – weareallin Feb 10 '19 at 5:12
• But I said that the length of $BP$ is $\sqrt350$, not the side of the square $ADCP$. – Anirban Niloy Feb 10 '19 at 5:24
• I thought you said that the area of the square is $350$? Oh, I see now, you didn't include the larger square in your diagram. Sorry, my solution is incorrect. – weareallin Feb 10 '19 at 5:46