# Compute numerically the angle $x$ in the triangle without trigonometry

Be $$\triangle CAB$$ right in $$A$$ such that $$AB=a$$ and $$\angle CBA = \alpha$$.

Extend $$BC$$ to $$D$$ such that $$\angle CAD=2\alpha$$ and $$\angle ADC=x$$.

If $$M$$ is a point $$\in BC$$ such that $$BM=MC$$ and $$MD=a$$, then compute $$\angle ADC=x$$

I tried drawing the altitude $$h$$ from $$A$$ to $$BC$$ such that $$AH=h$$, and then i tried some relations between similar triangles, but found nothing.

Any hints?

PS: If anyone has a solution that involves trigonometry, i would like to see it, but i highly prefer a solution that doesn't involve trigonometry.

• Angle $BCA$ is $90-\alpha$, angle at $ACD$ is $90+\alpha$, so $x=90-3\alpha$. – abiessu Mar 12 at 5:02
• Well, i mean a numerical answer, that's why i put "compute" instead of "compute in terms of $\alpha$" because that's trivial. I'm going to edit my answer to clarify anyways. – Rodrigo Pizarro Mar 12 at 5:10

Naturally, the first step is to extend the diagram. We have $$AB=MD$$; let's take advantage of that by constructing $$E$$ below $$BD$$ so that triangles $$MED$$ and $$ACB$$ are congruent.

From the right angle at $$A$$, the midpoint of $$BC$$ is the circumcenter of $$ABC$$, and $$AM=BM=CM$$. Similarly, if we mark the midpoint $$F$$ of $$DE$$, $$DF=EF=MF$$. Since $$BC=DE$$, all six of these lengths are equal; call that length $$b$$.

Now, it's time to use the angle marked in red. We also have $$\angle AMC=2\alpha$$ from the isosceles triangle $$BAM$$, and that makes triangles $$MAD$$ and $$ACD$$ similar. In particular, $$\angle MAD=\angle ACD=90^\circ+\alpha$$. Combine that with $$\angle MED = 90^\circ-\alpha$$, and $$EDAM$$ is a cyclic quadrilateral. From our earlier note that $$DF=EF=MF$$, then, $$F$$ is the circumcenter of $$EDAM$$.

So now, we've marked all of the segments of length $$y$$ in purple. Of particular note are the newly drawn segments $$AM$$, $$FM$$, and $$AF$$. That's an equilateral triangle $$FAM$$. On top of that, we know angles $$AMD$$ and $$DMF$$, so we can deduce that $$60^\circ=\angle AMF=3\alpha$$.

We were asked to find $$\angle CDA$$. From the angles of $$2\alpha$$ at $$A$$ and $$90^\circ+\alpha$$ at $$C$$, this third angle of the triangle is $$90^\circ-3\alpha=90^\circ-60^\circ=30^\circ$$. Done.

Refer to the diagram:

From $$\triangle ABC$$: $$\cos \alpha =\frac{y}{a} \Rightarrow a=2y\cos \alpha$$.

From $$\triangle AMD$$: $$\frac{a}{\sin (90^\circ +\alpha)}=\frac{y}{\sin x} \Rightarrow a\sin x=y\cos \alpha$$.

Can you find $$x$$ now?

It is:

Dividing the equations results in: $$\sin x=\frac12 \Rightarrow x=30^\circ.$$