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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

enter image description here drawing is not to scale...

Figure mentioned by colleague Ivan:

enter image description here

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  • $\begingroup$ Do we know that angel of $APH$? Is that angel colored orange because it is $53^\circ$? If so, first solve for $HP$ and then for $HC$. $\endgroup$
    – Roland Killian
    Commented Oct 13, 2021 at 17:33
  • $\begingroup$ @ACB I also found it...but I only post the question as an exact copy of the book...I For me, $H \in AC$ and $AH = 12\\PQ \perp AC, Q \in AB, ~and ~P \in PB ~and ~AP=AB $. $\endgroup$
    – peta arantes
    Commented Oct 13, 2021 at 18:53
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    $\begingroup$ But what about angles? You write $PAB=53^\circ, \angle 143^\circ$. Which angle is $53$ and which is $143$ as per the question? $\endgroup$
    – Math Lover
    Commented Oct 14, 2021 at 0:25
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    $\begingroup$ @MathLover forgive my distraction...I've already made the correction..$\angle PAB = 53^o$, ~ $\angle ACB = 143^o$ and redid the figure $\endgroup$
    – peta arantes
    Commented Oct 14, 2021 at 10:29
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    $\begingroup$ @MathLover...I think Ivan unlocked the "secret"..see the new drawing..I think it is now possible to solve by geometry $\endgroup$
    – peta arantes
    Commented Oct 15, 2021 at 12:32

2 Answers 2

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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.

IMG

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, $\small HC=8$.

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  • $\begingroup$ The answer is not precise 8, because tan(63.5°) is not precise 2. The solution is not only by geometry, because you use trigonometrical fact tan(63.5°)≈2. $\endgroup$
    – Ivan Kaznacheyeu
    Commented Oct 15, 2021 at 14:38
  • $\begingroup$ Yes, exactly. But as I know, the book that OP is referring to, always considers special right triangles as in my answer. Have a look at OP's past questions : ) E.g. math.stackexchange.com/q/4234366/947379 $\endgroup$
    – ACB
    Commented Oct 15, 2021 at 14:39
  • $\begingroup$ Understood, but I think it is something wrong in avoiding trigonometry by using such tricks, that are still based on trigonometry. $\endgroup$
    – Ivan Kaznacheyeu
    Commented Oct 15, 2021 at 14:42
  • $\begingroup$ @ACB excellent...I'm glad Ivan realized where the error was. $\endgroup$
    – peta arantes
    Commented Oct 15, 2021 at 15:25
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The answer is not precise 8.

If you make drawing placing P up to AC.

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

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  • $\begingroup$ very good..I think you've unlocked the "secret" $\endgroup$
    – peta arantes
    Commented Oct 15, 2021 at 12:30
  • $\begingroup$ would it be possible to have a resolution without trigonometry? .I imagine that maybe because of remarkable triangles depending on the angles that appear $\endgroup$
    – peta arantes
    Commented Oct 15, 2021 at 12:54
  • $\begingroup$ Aha, makes sense now that I read it! $\endgroup$
    – Math Lover
    Commented Oct 15, 2021 at 13:32

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