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?
-
0However, if you'd rather the OP improve the drawing (redraw if necessary), I'm fine with that. – 2011-05-17
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.
- Label the point of intersection of $\overline{AE}$ and $\overline{CD}$ as $X$.
- 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 $CX=XD$.
- From $m\angle ADC=m\angle BAE$, $\triangle ADX$ is isosceles with $AX=DX$.
- So we have $AX=DX=CX$.
- 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 $D$.
- $\overline{CD}$ is a diameter of the circle, since it contains the center $X$.
- $\angle DAC$ 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 $\angle BAC$.
-
0can you talk details about the circle?....i need another way to do this problem – 2011-05-18
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.
-
0@Briana791: There would be several converses. For example, suppose angle at $A$ is $90$ degrees, and $\angle ADC=\angle BAE$ and $D$ is midpoint of $AB$. Then it follows that $BE=2EC$. About the previous question, I don't know what you mean. And about other proofs (you asked earlier), yes, there are, but they are less "elementary" so if you are having so much trouble with this one, it doesn't seem worthwhile to describe alternatives. – 2011-05-18