How to prove an angle equals to $90^°$ algebraicly Let $ABCD$ be an isosceles trapezoid. Let $P$ be the intersection of its diagonals. the line $DA$ meets the circumcircle of $APB$ at $M$. Let $L$ be a point on $AB$ such that $LD=LB$ and $K$ be a point on $DL$ such that $KLAM$ is cyclic. Denote by $J$ the intersection of $KA$ and $ML$. 
Prove that $\angle BPJ =90^°$ 
Sorry I haven't tried to solve this, it's really hard but I did a geometric interpretation and the angle is right. 
 A: Thank you @pppl234plp for forcing me to try out this wonderful problem. Here goes the solution :-
We avoid directed angles and stick with this configuration.
Firstly note that $DA\times DM=DL \times DK=DP \times DB$ so that by converse of the Power of Point, we get $LPBK$ is cyclic.
Next, $\angle LBP=\angle LKP =\angle LDP$ so that $PK=PD=PC$ where the last equality comes from the fact that $ABCD$ is an isosceles trapezoid.
Next we claim that $ALPD$ is cyclic.
Proof:Note that $\angle LDP=\angle LBP=\angle ABP=\angle BAP$ (isosceles trapezoid) $=\angle LAP$ and the claim follows.
We now claim that $DK=DC$.
Notice that $\angle KPD=180-\angle KPB=180-\angle KLB=180-\angle ALD=180-\angle APD=\angle DPC$.
Now, $KP=KC$, segment $PD$ is common and $\angle KPD=\angle DPC$ so that triangles $KDP$ and $PDC$ are congruent, and hence $KD=DC$ and the claim follows.
Note now that $\angle BKL=\angle LPD=\angle MAL$ so that $M,K,B$ are collinear.
This readily gives $P$ is the miquel point of the complete quadrialteral $LAMKDB$
As we have the complete quadrilateral $LAMKDB$ and as $LAMK$ is cyclic hence, $J$ and $P$ are swap each other under the inversion $\Psi$ centered at the center of $\odot (MKAL)$ and radius the same as of $\odot (MKAL)$, so that if we let $O$ denote center of $\odot (MKAL)$ then $O,J,P$ are collinear.
As $DB$ is the polar of $J$ with respect to $\odot (MKAL)$ (Brokard's Theorem), line $OJ=$ line $JP$ is orthogonal to $BD$ and the conclusion follows.
