How to find the angle formed by a cevian in a right triangle given a condition?

Question

The problem is as follows:

In a right triangle $\triangle ABC$, right at $B$, it is traced the height $BH$, in the triangle $BHC$, then is traced the interior cevian $HM$ in such a way that $MC = AB$. Find the measure of the angle $\angle MHC$ if it holds that $HC = BH + 2AH$.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&\frac{53^{\circ}}{2}\\ 2.&\frac{37^{\circ}}{2}\\ 3.&53^{\circ}\\ 4.&37^{\circ}\\ 4.&30^{\circ}\\ \end{array}$

In this particular problem. I'm stuck at the very beginning. What's exactly the right drawing or interpretation for this?. Can someone help me here?. Can this be solved using congruency?. Which among the identities of congruency of triangles would apply?.

Please include a drawing in the answer because I don't know which sort of appropiate drawing would be for this particular problem. Can it be solved using only euclidean geometry postulates?.

Typically I would put or place an effort in the solution of this problem, but in this particular subject I'm stuck at the very beginning because I don't know how to translate what it is written in words into the problem with accuracy so that the problem can be solved, thus I really need help.

If it helps, I found this problem in a puzzles book which goes by the name Reason and logic which is from 2000s and includes topics probably borrowed from Martin Gardner's 50's book and Moise Geometry's textbook from the 70's.

I really wish this can suffice to give information. So far, as I indicated in the problem my best guess is that the problem uses triangle congruency, but I don't know how to make the right sketch for making this to happen.

Upon on my progress I could spot that there are going to be formed two congruent triangles and if this condition is used then I believe the problem can be solved, hence what might be needed is to establish a relationship between the altitude from $M$ to $HC$ would allow the answer. Such approach is mentioned in the answer given below.

Therefore, since I'm not good in this topic, it would help me a lot if the answer would include a step by step solution so I can understand.

Answer 1

The interpretation of the problem is as follows:

Please see the diagram from below:

This first diagram is a representation of what its intended to say in the problem.

The key part here is that $MC=AB$ this will be used later in the problem.

The following diagram is the solution and the way to solve it is by tracing a line as the altitude from $MP$. In order to take track of the congruency. Angles $\alpha$, $\beta$ and $\omega$ as auxiliary are used.

Therefore the angle can be obtained from the triangle $\triangle HPM$.

I noted that there's congruence of triangles: This is the part where its used the information given in the problem regarding $AB=MC$.

$\triangle PMC \cong \triangle AHB$ because of $A-S-A$. For additional information and reference this link can explain the case.

This means $PC=BH$ and $AH=PM$

Here's where we use the second clue given in the problem where:

$HC=BH+2AH$

Since:

$HP = HC-PC$ and $PC=BH$

$HP = BH+2AH-BH = 2AH$

and since $PM=AH$ we're left with a right triangle whose sides are in a proportion of $1:2$.

This means that:

$\tan \omega = \frac{AH}{2AH}=\frac{1}{2}$

$\angle MHC = \frac{53^{\circ}}{2}$

Without resorting to that trigonometric function it can easily be inferred that the intended triangle corresponds to:

The ratio of:

$1-2-\sqrt{5}$

is none other than $\frac{53^{\circ}}{2}$.

Answer 2

By the chord-chord power theorem with respect to the point $H$, \begin{align} |BH|^2&=|AH|\cdot|CH| \tag{1}\label{1} . \end{align}

Given that

\begin{align} |CH|&=|BH|+2\cdot|AH| \tag{2}\label{2} , \end{align}

we have a quadratic equation

\begin{align} \left(\frac{|BH|}{|AH|}\right)^2 &= \frac{|BH|}{|AH|}+2 \tag{3}\label{3} , \end{align}

hence \begin{align} \frac{|BH|}{|AH|} &=2=\tan\alpha \tag{4}\label{4} ,\\ \cos\alpha&=\tfrac{\sqrt5}5 ,\quad \sin\alpha =\tfrac{2\sqrt5}5 \tag{5}\label{5} . \end{align}

Let $|OA|=|OC|=R=\tfrac12$. Then

\begin{align} c&=|AB|=|CM|=\cos\alpha=\tfrac{\sqrt5}5 \tag{6}\label{6} ,\\ |AH|=|MP|&=c\cos\alpha=\cos^2\alpha =\tfrac15 \tag{7}\label{7} ,\\ |BH|=|CP| &=|AH|\tan\alpha =\tfrac25 \tag{8}\label{8} ,\\ |CH|&=\tfrac45 \tag{9}\label{9} ,\\ |HP|&= |CH|-|CP|=\tfrac25 \tag{10}\label{10} ,\\ \tan x&=\frac{|MP|}{|HP|} =\tfrac12 \tag{11}\label{11} ,\\ x&=\arctan \tfrac12 \approx 26.565^\circ \tag{12}\label{12} . \end{align}