0
$\begingroup$

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

$\endgroup$
  • 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
5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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: $\sin x=\frac12 \Rightarrow x=30^\circ.$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.