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?

In the given figure C is the midpoint of AB,D is the midpoint of XY and AC=XD Using an Euclid \\’…
In the given figure C is the midpoint of AB,D is the midpoint of XY and AC=XD Using an Euclid \\’…

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

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?

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

  • 1$\begingroup$ Have you tried making a drawing? $\endgroup$ May 16, 2011 at 16:39
  • 2$\begingroup$ I think user forgot to include the measure of angle BAE… @Briana: could you edit your post, and this time include the question in the post itself, along with missing info? $\endgroup$ May 16, 2011 at 16:39
  • 1$\begingroup$ @Briana791: Please edit your post, being careful to indicate carefully what the problem is. There are some puzzling things about your question. For example, you mention that $D$ is the midpoint of $AB$. But $D$ seems to have nothing further to do with the question. And you write “If $m\angle BAE$.” If what is known about this angle? Don’t try to squeeze it all into the title! $\endgroup$ May 16, 2011 at 16:41
  • 5$\begingroup$ I’d say that if the OP doesn’t complete the question, perhaps it should be closed? $\endgroup$ May 16, 2011 at 17:19
  • 5$\begingroup$ I voted to reopen, but I’d recommend that you explain a bit what you’ve tried. Also, the scan is a bit hard to read. (please vote this comment up so that it can be seen) $\endgroup$– t.b.May 17, 2011 at 15:51

2 Answers

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 $\angle BAC$.
  • $\begingroup$ im sorry but this is confusing for me can u be detailed? $\endgroup$ May 17, 2011 at 17:01
  • $\begingroup$ @Briana791: I’d be happy to be more detailed, but I need to know more about where the problem came from (what else you’ve been doing around/near this that might be related) and/or more about what, if any, of my solution you did understand. $\endgroup$– IsaacMay 17, 2011 at 17:04
  • $\begingroup$ this problem came from a calendar problem from NCTM site. its the third set on spring 2011. i understand that part where u made the intesection point X. but i dont get the ratio part, how AX and CX, and DX are equal, and the circle part. $\endgroup$ May 17, 2011 at 17:10
  • 2$\begingroup$ @Briana: you mentioned when you first post this question that you needed to include it in your “portfolio”…the fact that this problem is from the NCTM site (Teachers of mathematics), I’m wondering if you need this for your portfolio is related to teacher certification to demonstrate problem solving skill. If so, you need to try to make connections between what is given, and what conclusions you can start with…and be more specific about where you get hung up. $\endgroup$ May 17, 2011 at 17:18
  • $\begingroup$ its for my problem solving class my professor gets the problems from the website $\endgroup$ May 17, 2011 at 17:28

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.

  • $\begingroup$ so what would be the answer of ∠BAC? $\endgroup$ May 17, 2011 at 18:49
  • $\begingroup$ Ah, very nice! I had a feeling that working around with areas might get there, but mass points was staring me in the face and I was having trouble seeing past it. $\endgroup$– IsaacMay 17, 2011 at 19:20
  • $\begingroup$ @Briana791: If you follow the diagram, you will see that $\angle BAC$ is what I called $p+q$, so $90$ degrees. Or else use the fact that the angle subtended by the diameter of a circle is $90$ degrees. $\endgroup$ May 17, 2011 at 19:25
  • $\begingroup$ but it doesnt look like 90 degrees $\endgroup$ May 17, 2011 at 19:37
  • 1$\begingroup$ @Briana791: The picture is somewhat imperfect. And it kind of looks like $\angle BAC$ is not far from $90$ degrees. Anyway, it is $90$. A careful picture using good drawing software would show it. Maybe it was drawn a little off so you would not guess correctly. $\endgroup$ May 17, 2011 at 19:54

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