If the area of $ ABP$ is $ 192 $ find $ PA*PC $ Let $ABCD$ be an isosceles trapezium with bases $ AB=32 $ and $CD=18$.
Inside $ABCD$ there's a point $P$ such that $ \angle PAD= \angle PBA $ and $ \angle PDA =\angle PCD $.
If the area of $ ABP$ is $ 192 $ find $ PA*PC $.
My try: 
lead by $P$ perpendicular to the basics; it intersects the base points $E$ and $F$. $S_{ABP}=\frac12AB\cdot PE$ $\Rightarrow$ $PE=12$. How prove $AE=EB$?

 A: The answer is $PA$ * $PC$ = $20$*$15$ = $300$.
As maxkor showed, the perpendicular distance from the point $P$ to the base at point $E$ is $12$. 
Let's suppose that $AE$ = $EB$, as maxkor suggested. Assuming origo is at point $A$, this means the point $P$ is located at $(16,12)$. Let us further suppose that the height of the trapezium is $24$.
This means the vertices of the trapezium are located as follows:
$A = (0, 0)$
$B = (32, 0)$
$C = (25, 24)$
$D = (7, 24)$
The calculation of $PA$ and $PC$ is now trivial given the coordinates of all vertices and the point $P$. However, we need first to show that this trapezium fulfills the requirements, i.e. that $\angle PAD= \angle PBA$ and $\angle PDA =\angle PCD$. We can do this using the Law of Cosines:
$Cos(\angle PAD) = \frac{15^2-(25^2+20^2)}{-2*25*20} = 0.8$
and 
$Cos(\angle PBA) = \frac{12^2-(16^2+20^2)}{-2*16*20} = 0.8$
Hence $\angle PAD= \angle PBA$.
$Cos(\angle PDA) = \frac{20^2-(25^2+15^2)}{-2*25*15} = 0.6$
and 
$Cos(\angle PCD) = \frac{15^2-(18^2+15^2)}{-2*18*15} = 0.6$
Hence $\angle PDA= \angle PCD$.
QED.
A: Draw the circle through $PAB$: from $\angle PAD= \angle PBA$ it follows that $AD$ and $BC$ are tangent to the circle at $A$ and $B$. In the same way, the circle through $PCD$ is tangent to $AD$ and $BC$ at $D$ and $C$.
It is well known, on the other hand, that the radical axis $PQ$ bisects the common tangents $AD$ and $BC$ of the two circles, because $KP\cdot KQ=CK^2=BK^2$. That entails $PF=PE=12$, $DH=24$ and $BC=AD=\sqrt{24^2+7^2}=25$.
From the similitude of triangles $PKC$ and $CPD$ one gets $CP:CD=PK:PC$, that is: $PC^2=18PK$. Analogously, from the similitude of triangles $JPA$ and $APB$ one gets $PA:AB=PJ:PA$, that is: $PA^2=32PJ$.
It follows that: 
$$
PA^2\cdot PC^2=18\cdot32\,PJ\cdot PK=18\cdot32\, QK\cdot PK=
18\cdot32\, CK^2,
$$
so that
$$
PA\cdot PC= 72\, CK= 24\,{BC\over2}=300.
$$

