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In figure $\triangle APB$ , $\triangle QBC$ & $\triangle ARC$ are isosceles. $PK$ $\perp AC$, also $Ql$ $\perp AC$ $P,B,L$ are co linear. Also $\angle APB =\angle BQC = \angle ARC = 120^o$. How to prove by symmetry that $\angle KPQ = \angle PKL$. Please give elaborated prove this. enter image description here

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$PK$ and $QL$ are parallel since both are perpendicular to $AC$. Let the point where $PK$ intersects $AC$ be called $T$. Because $\angle PAT = \angle KAT = 30^o$, $PT = TK$ and it follows that $\triangle PAT \cong \triangle KAT$. By the SAS property, $\triangle PBT \cong \triangle KBT$, and therefore $\angle PBT = \angle KBT=30^o$. Because $\angle QBC = 30^o$, $\angle PBQ = 120^o$. It follows from here $K$, $B$ and $Q$ are colinear. Consequently, it may now be stated that $KQ$ and $PL$ are intersecting diagonals of the trapezium $PQLK$. Because $AC$ bisects $QL$ and $PK$ at right angles, the trapezium $PQLK$ is an isosceles trapezium.

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First of all, we you must prove that $P,B,L$ are collinear, this is not given, it follows from the other given information.

For the equality of angles.

Since $PK\bot AC$ and $\angle PAC = \angle KAC = 30^{\circ}$ it follows that reflection across $\ell = AC$ maps $P\mapsto K$ and vice versa. The same is true for $Q\mapsto L$ and vice versa. So this reflection maps $$\angle LKP \mapsto \angle QPK $$ and thus they are of equal size.

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