<s>[INST] In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm. 
In trapezoid $ABCD$, $AB \parallel CD$ , $AB = 4$ cm and $CD = 10$ cm. Suppose the lines $AD$ and $BC$ intersect at right angles and the lines $AC$ and $BD$ when extended at point $Q$ form an angle of $45^\circ$. Compute the area of $ABCD$.

What I Tried :- Here is the picture :-
Now to find the area of $ABCD$, I just need to find its height, but I cannot find it.
I can see that $\Delta AOB \sim \Delta COD$. So :-
$$\frac{AB}{CD} = \frac{AO}{OD} = \frac{BO}{OC} = \frac{2}{5}$$
So I assumed $AO = 2x$ , $BO = 2y$ , $CO = 5y$ , $DO = 5x$. 
Now in $\Delta AOB$, by Pythagorean Theorem :- 
$AO^2 + OB^2 = AB^2$ 
$\rightarrow 4x^2 + 4y^2 = 16$ 
$\rightarrow x^2 + y^2 = 4$
Also $\Delta QAB \sim \Delta QDC$. So:-
$$\frac{QA}{AC} = \frac{QB}{BD}$$ 
I get $AC$ and $BD$ by Pythagorean Theorem again, which gives me :-
$$\frac{QA}{\sqrt{4x^2 + 25y^2}} = \frac{QB}{\sqrt{25x^2 + 4y^2}}$$
I don't know how to proceed next, as this result only gives me that $\left(\frac{QA}{OB}\right)^2 = \frac{21y^2 + 16}{21x^2 + 16}$ . Also I couldn't think of any way to use the $45^\circ$ angle, except that I can figure out that the triangle is cyclic.
Can anyone help?
 [/INST] By your work $$(2x)^2+(2y)^2=4^2,$$ which gives $$x^2+y^2=4.$$
Also, $$AD=\sqrt{DO^2+AO^2}=\sqrt{25y^2+4x^2}$$ and
$$BC=\sqrt{25x^2+4y^2}.$$
Now, let $PABC$ be parallelogram.
Thus, $P\in DC$, $AP=BC$, $$DP=DC-PC=10-4=6$$ and $$\measuredangle DAP=\measuredangle Q=45^{\circ}.$$ Thus, by the law of cosines for $\Delta DAP$ we obtain:
$$\frac{25y^2+4x^2+25x^2+4y^2-36}{2\sqrt{(25x^2+4y^2)(25y^2+4x^2)}}=\frac{1}{\sqrt2}$$ or
$$\frac{29\cdot4-36}{\sqrt2}=\sqrt{(25x^2+4y^2)(25y^2+4x^2)}$$ or
$$3200=641x^2y^2+100(x^4+y^4)$$ or
$$3200=441x^2y^2+100(x^2+y^2)^2,$$ which gives
$$xy=\frac{40}{21}.$$
Id est, $$S_{ABCD}=\frac{1}{2}AC\cdot DB=\frac{1}{2}\cdot7x\cdot7y=\frac{140}{3}.$$
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