Question: 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 $AE=EB$?

[![enter image description here](https://i.stack.imgur.com/x1Nrf.png)](https://i.stack.imgur.com/x1Nrf.png)

Answer: 
Suppose $f$ is not continuous from the right at $x\_0$ and $f(x\_0)=y\_0$. Notice that for all $x>x\_0$, $f([x\_0,x])$ is compact. Then there exists a sequence $\{x\_n\}\_{n\in\mathbb{N}}$ and $y\_1\neq y\_0$ such that $x\_n>x\_0$, $\lim\_{n\to+\infty}x\_n=x\_0$ and $\lim\_{n\to+\infty}f(x\_n)=y\_1$. Furthermore, $y\_1\in f([x\_0,x])$ for all $x$ because of the compactness. Also $f^{-1}(\{y\_1\})$ is closed, and $f(x\_0)=y\_0\neq y\_1$. So $x\_0$ lies in an open set $(a,b)$ such that $y\_1\notin f((a,b))$. It's a contradiction.
Hint: By the Heine-Borel theorem, every compact subset of $\mathbb R$ is closed and bounded. Therefore, f carries closed sets into closed sets.

Claim: A function f from X into Y is continuous if and only if preimages of closed sets in Y are closed in X.

If you can prove this,this will show the answer is yes.