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.

enter image description here

  • 1
    $\begingroup$ Angle $BCA$ is $90-\alpha$, angle at $ACD$ is $90+\alpha$, so $x=90-3\alpha$. $\endgroup$ – abiessu Mar 12 at 5:02
  • $\begingroup$ 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. $\endgroup$ – 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.

Figure 1

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 $EDAM$.

Figure 2

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:

enter image description here 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: $\sin x=\frac12 \Rightarrow x=30^\circ.$


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