What's the measure of the segment HC in the triangle below?

Question

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:

Answer 1

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, $\small HC=8$.

Answer 2

The answer is not precise 8.

If you make drawing placing P up to AC.

$AH=AP \cos (53^o+\angle BAC) = AB \sin(37^o-\angle BAC)=AB \sin \angle ABC=AC \sin 143^o \Rightarrow AC=AH \csc 143^o \Rightarrow HC=AH(\csc 143^o-1)\approx 8$.