In △ ABC, D is the midpoint of AB, while E lies on BC satisfying BE = 2EC. If m∠ADC=m∠BAE, what is the measure of ∠BAC in degrees?

Question

In △ABC, D is the midpoint of AB, while E lies on BC satisfying BE = 2EC. If m∠ADC=m∠BAE, what is the measure of ∠BAC in degrees? I know already that angle A and angle D are congruent because m∠ADC=m∠BAE I just dont know how to approach the problem and i would love some help

Answer 1

edit Given that this was an NCTM calendar problem, I doubt that my solution is the intended one, but I've added some more detail to flesh it out.

1. Label the point of intersection of $\overline{AE}$ and $\overline{CD}$ as $X$.
2. Use the technique of mass points:
   • Since $D$ is the midpoint of $\overline{AB}$, $AD:DB=1:1$.
   • Since $BE=2\cdot EC$, $BE:EC=1:2$.
   • Putting masses of $1$ at $A$, $1$ at $B$ and $2$ at $C$ will allow the triangle to balance at $X$.
   • The "effective mass" at $D$ is $2$ (the sum of the masses at $A$ and $B$).
   • The ratio $CX:XD$ is equal to the ratio of the masses $D:C$, so $CX:XD=2:2=1:1$, or $CX=XD$.
3. From $m\angle ADC=m\angle BAE$, $\triangle ADX$ is isosceles with $AX=DX$.
4. So we have $AX=DX=CX$.
5. So $A$, $C$, and $D$ are all the same distance from $X$. A Circle is the set of points that are a fixed distance from a fixed point. So there is a circle with center $X$ that contains $A$, $C$, and $D$.
6. $\overline{CD}$ is a diameter of the circle, since it contains the center $X$.
7. $\angle DAC$ is inscribed in the circle, or, since $\overline{CD}$ is a diameter, $\angle DAC$ is inscribed in a semicircle. From this, you can conclude the measure of $\angle DAC$, which is the same as the measure of $\angle BAC$.

Answer 2

This solution will be essentially the same as the one given by @Isaac, except for the appeal to Ceva's Theorem. While reading the words, please look at a diagram: the whole thing would be much easier to explain at a blackboard by pointing!

I have tried to use only a very basic geometric fact about area of a triangle, that it is half of base times height.

Let $X$ be the point where lines $CD$ and $AE$ meet. Draw the line $BX$, and let this meet side $CA$ at $F$ (which will never be mentioned again, but I like symmetry).

Note that the area of $\triangle AEB$ is twice the area of triangle $ACE$ (the base $EB$ of $\triangle AEB$ is twice the base $CE$ of $\triangle ACE$, and the heights are the same).

For the same reason, the area of $\triangle XEB$ is twice the area of $\triangle XCE$.

It follows by subtraction that the area of $\triangle AXB$ is twice the area of $\triangle ACX$.

But the area of $\triangle AXB$ is twice the area of $\triangle AXD$.

We conclude (this is the important conclusion) that the area of $\triangle ACX$ is equal to the area of $\triangle AXD$.

But $\triangle AXD$ has base $XD$, and $\triangle ACX$ has base $CX$. With respect to these bases, the two triangles have the same height. Since their areas are equal, their bases are equal.

We conclude that $CX=XD$. And, by what we were given, each of these is equal to $XA$.

Now we can, as suggested by @Isaac, draw the circle with center $X$ and radius $XD$. This circle passes through $A$. Now use a basic fact about angle subtended by a diameter.

Or else do some angle-chasing. We have shown that $XC=XA$. Let $\angle ACD$ be $p$ (degrees). Then $\angle CAX=p$.

Also, $\triangle AXD$ is isosceles. Let $\angle XAD=\angle XDA=q$.

Then the angles of $\triangle ACD$ add up to $2p+2q$. But they add up to $180^\circ$. Thus $p+q=90^\circ$.

Please remember that this is the solution of @Isaac, so if you wish to accept, his is the right one to accept.