Can Isosceles Triangles Solve This Geometry Problem?

[karush]

View attachment 5803
I tried about an hour to solve this but couldn't get the ratios to work I assume there are isoseles triangles in this but that is just observation

 

[Opalg]

I do not believe that there is enough information here to provide a solution. You need an additional fact, such as for example the length of the sides of the square.

The point $P$ must lie on an Apollonius circle with its centre on the line $BD$. (That is the set of all points for which the distances $PD$ and $PB$ are in the ratio 5 to 6.)

The area of the triangle $APC$ is half the diagonal $AC$ times the distance of $P$ from $AC$. But unless you know the length of that diagonal you cannot fix the position of $P$ or the lengths $PB$ and $PD$.

Edit. ILS's neat solution below shows that I was wrong. I was trying to determine the position of the point $P$, and it's true that this cannot be determined without further information. But the lengths $PB$ and $PD$ are uniquely determined, which I find quite surprising.

 

[I like Serena]

Sorry Opalg, I'm afraid I have to disagree. (Wink)

Spoiler

Let's say the square has side $s$.
Let's pick $A$ at the origin $O$.
And lets' pick $P=(p,q)$.
Then $A=(0,0), B=(s,0), C=(s,s), D=(0,s)$.

The triangle $ACP$ has area:
$$\frac 12 \| \vec{AP} \times \vec{AC} \| = \frac 12 \| \vec{OP} \times \vec{OC} \| = \frac 12 |ps-qs| = 19 \tag 1$$
The other equations yield:
$$PD=15x \quad\Rightarrow\quad p^2 + (s-q)^2 = (15x)^2 \quad\Rightarrow\quad p^2+q^2+s^2 - 2qs = 225x^2 \tag 2$$
$$PB=18x \quad\Rightarrow\quad (s-p)^2 + q^2 = (18x)^2 \quad\Rightarrow\quad p^2+q^2+s^2 - 2ps = 324x^2 \tag 3$$

Subtract $(2)$ from $(3)$:
$$-2ps + 2qs = 99x^2$$

Combine with $(1)$:
$$4 \cdot 19 = 99x^2 \quad\Rightarrow\quad x = \sqrt{\frac{4\cdot 19}{99}} = \frac 23\sqrt{\frac{19}{11}}$$

 

[karush]

wow thank you

I spent another 2 hours on this but couldn't make those conections

I found this on G+ but everyone was giving up. but here it got solved.