For reference: In triangle $ABC$, in $AC$ the point $H$ is considered.
By $H$, the perpendicular $PH$ to $AC$ is drawn which intersects $AB$ at $Q$.
$PAB=53^o, \angle ACB =143^o$
$AP=AB, AH=12$
Calculate HC. (Answer:8)
My progress…
I think this is the drawing…I identified a menelaus in the triangle $ABC-QP \implies BQ.12.CP = QA.HC.PB$
but it didn't help much…
$\triangle PAB(isosceles) PA = PB \implies \angle P = \angle B$
Through geogebra I couldn't make the drawings according to the data provided… if I follow the data and the answer, the angles will not be the same,,, either I drew wrong or the problem is in the statement
drawing is not to scale…
Figure mentioned by colleague Ivan:
Best Answer
Many thanks to @Ivan for identifying the error in the original image, and here is a solution by geometry as per the request of OP.
Extend $\small PH$ to meet $\small BC$ extend at $\small G$. $\angle CGH=53^\circ$.
$\small \therefore\angle BGP=\angle BAP\implies AGBP\ \text{is cyclic}$
As $\small AB=AP$, $\small \angle ABP=63.5^\circ$, and so is $\small \angle AGP$ because they are in the same segment.
Therefore, $\small \triangle AHG$ is a special right triangle with perpendicular sides in the ratio 1:2.
$\small \therefore HG=6$
Similarly, as $\small \triangle CGH$ is a $\small 3:4:5$ right triangle, $\small HC=8$.