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## Isosceles Problems

One of my students gave me a problem last fall that was very interesting.

The problem is posed with the following diagram:

What is the measure of $\angle CDA$?  That is, what is the value of $x$?

I’ve given a similar problem in my trigonometry class for the past few years, except that version of the problem has a side length included and the triangle is not isosceles.

A pdf of this problem is linked below:

jun_7_mth_112_river_problem

Working from my experience with the other version of this problem, I began to write in values for the various unlabeled angles in the diagram – if we label the intersection of $\overline{AD}$ and $\overline{BC}$ as $K$, then $\angle CKD$ and $\angle AKB$ are both $70^{\circ}$, $\angle CKA$ and $\angle DKB$ are both $110^{\circ}$, which makes $\angle KCA$ $50^{\circ}$ and $\angle ADB$ is $40^{\circ}$.

I added in new variables and created a system of four equations with four unknowns, but it was a dependent system.

The solution for this problem that was devised by the student who gave it to me is after the jump…

His solution was primarily synthetic and not algebraic!

His first step was to draw line segment $\overline{AE}$ such that $\overline{AE}$ and $\overline{AB}$ are congruent, creating a small isosceles triangle in the interior of the larger one.  Then $\angle ABE$ and $\angle AEB$ are both $80^{\circ}$ and $\angle EAB$ is $20^{\circ}$

Once this is done, we can add back in $\overline{BC}$ and look at triangle $\triangle ABC$.  Notice that $\angle ABC=50^{\circ}$ and $\angle CAB=80^{\circ}$, so that $\angle BCA$ is also $50^{\circ}$, making $\triangle ABC$ isosceles with sides $\overline{AB}$ and $\overline{AC}$ congruent.

Now, if we draw $\overline{EC}$, we know that $\overline{AB}$, $\overline{AE}$ and $\overline{AC}$ are all congruent.  With $\overline{AE}$ and $\overline{AC}$ congruent, then $\triangle AEC$ is isosceles.  With vertex angle $\angle CAE=60^{\circ}$, then both $\angle AEC$ and $\angle ACE$ are $60^{\circ}$ as well, making $\triangle AEC$ equilateral with side $\overline{EC}$ congruent to $\overline{AB}$, $\overline{AE}$ and $\overline{AC}$.

Then, we add $\overline{AD}$ back into the diagram, $\angle DAE$ is now $40^{\circ}$ as the original angle $\angle DAB$ was $60^{\circ}$ and $\angle EAB$ is $20^{\circ}$.

So, with $\overline{AD}$ redrawn and $\angle DAE=40^{\circ}$, then in triangle $\triangle ADE$, $\angle DAE=40^{\circ}$ and $\angle DEA=100^{\circ}$, then $\angle EDA$ must also be $40^{\circ}$, making $\triangle ADE$ isosceles and sides $\overline{ED}$ and $\overline{AE}$ congruent.

But side $\overline{AE}$ is already congruent to $\overline{EC}$ in the equilateral triangle $\triangle AEC$, so that means that $\overline{EC}$ and $\overline{ED}$ are congruent and $\triangle EDC$ is isosceles as well.  Once we know this, the solution is trivial – with $\angle CED=40^{\circ}$, then $140^{\circ}$ must be split equally between angles $\angle EDC$ and $\angle ECD$, so $\angle EDC=70^{\circ}$ and $x=30^{\circ}$!

The reasoning here is somewhat tricky and convoluted, but I looked for a simple algebraic solution to this for hours and got nowhere.

In researching this problem I did run across a similar but different problem.

Chris Harrow, Mathematics Chair at the Hawken School in Cleveland, OH, has a math blog called CAS Musings.  Here, he ponders a problem involving isosceles triangles with no angle measures given.  This problem generates a system of 6 equation with 6 unknowns that he solves using Wolfram Alpha.