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

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:

• 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 $HC$. Commented Oct 13, 2021 at 17:33
• @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 $AH = 12\\PQ \perp AC, Q \in AB, ~and ~P \in PB ~and ~AP=AB$. Commented Oct 13, 2021 at 18:53
• But what about angles? You write $PAB=53^\circ, \angle 143^\circ$. Which angle is $53$ and which is $143$ as per the question? Commented Oct 14, 2021 at 0:25
• @MathLover forgive my distraction...I've already made the correction..$\angle PAB = 53^o$, ~ $\angle ACB = 143^o$ and redid the figure Commented Oct 14, 2021 at 10:29
• @MathLover...I think Ivan unlocked the "secret"..see the new drawing..I think it is now possible to solve by geometry Commented Oct 15, 2021 at 12:32

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, $$\small HC=8$$.

• 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. Commented Oct 15, 2021 at 14:38
• 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
– ACB
Commented Oct 15, 2021 at 14:39
• Understood, but I think it is something wrong in avoiding trigonometry by using such tricks, that are still based on trigonometry. Commented Oct 15, 2021 at 14:42
• @ACB excellent...I'm glad Ivan realized where the error was. Commented Oct 15, 2021 at 15:25

The answer is not precise 8.

If you make drawing placing P up to AC.

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

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