# If one angle of two triangles and the two medians to the two sides forming this angle are equal are the two triangles congruent?

If one angle of two triangles and the two medians to the two sides forming this angle are equal, will the two triangles be congruent? For example, in triangles $$ABC$$ and $$A'B'C'$$, we have angle $$A = A'$$ and medians $$BD = B'D'$$ and $$CE = C'E'$$.

I have tried with Geogebra and indeed they are congruent. I used the fact that the 3 medians intersect at a point that has a distance of $$\frac{1}{3}$$ of its length from the origin vertex. By keeping one median fixed, I rotated the second one around the intersecting point and indeed there are two points which "see" the side $$BC$$ by a given angle, both of them resulting to the same triangles.

How do I solve it geometrically?

Let's name the intersection point of the 3 medians $$O$$. I started by assuming that $$BC \neq B'C'$$, then since the other two sides of the small triangles (formed by the $$\frac{2}{3}$$ of the medians) are equal, then angle $$BOC \neq B'O'C'$$. Then also angles $$DOE \neq D'O'E'$$, so, segment $$DE \neq D'E'$$. But I don't know how to go further.

Many thanks!

Consider the Figure above, where I used your notation, and at the beginning I followed a path similar to the one suggested by Michael in his answer.

1. Draw segment $$BD$$ (the first given median) and let $$O$$ be the centroid of the desired triangle, so that $$\overline{BO} = 2\overline{OD}$$.
2. Draw the unique (up to reflection with respect to $$BD$$) circumference $$\gamma_1$$ passing through $$B$$ and $$D$$ and such that the chord $$BD$$ subtends on it the desidered angle $$\angle BAD$$.
3. Draw then the circle $$\gamma_2$$ passing through the midpoint of $$BD$$, and internally tangent to $$\gamma_1$$ in $$B$$. This is the locus of the midpoints of any chord on $$\gamma_1$$ having one endpoint in $$B$$.
4. Draw $$\gamma_3$$, the circle centered in $$O$$ with radius equal to $$\frac13\overline{CE}$$, where $$CE$$ is the second given median.
5. $$\gamma_3$$ has at most one intersection point with $$\gamma_2$$ (in the half-plane where $$A$$, too, must lie).

Now it suffices to demonstrate that the above mentioned intersection point between $$\gamma_3$$ and $$\gamma_2$$ is in fact the uniquely determined point $$E$$.

In particular, since $$BE \cong AE$$ by our construction, we need to show that $$C = EO \cap AD$$ is such that $$AD \cong DC$$.

Consider triangle $$\triangle ABD$$ cut by $$EC$$. By Menelaus's Theorem you have $$\frac{\overline{BO}}{\overline{OD}}\cdot \frac{\overline{DC}}{\overline{AC}}\cdot \frac{\overline{AE}}{\overline{BE}} = 1,$$ hence $$\overline{AC} = 2\overline{DC},$$ and the thesis follows. $$\blacksquare$$

• dfnu: How do we construct (2)? Commented Nov 21, 2019 at 9:57
• @Samuel you mean the circle $\gamma_1$? Or $\gamma_2$?
– dfnu
Commented Nov 21, 2019 at 10:15
• @Samuel, if you are referring to $\gamma_1$, for example you can construct the unique (up to reflection) isosceles triangle $\triangle MBD$, with $\angle MBD \cong \angle MDB = \frac{\pi}2 - \angle BAC$. Then $\gamma_1$ has center $M$, and passes through $B$ and $D$. And of course $\gamma_2$ is the circle through $M$, $B$, and the midpoint of $BD$.
– dfnu
Commented Nov 21, 2019 at 10:23

Just a sketch to construct the triangle as follows. Draw $$BD$$ and determine $$O$$ such that $$BO:OD=2:1$$. Draw the chord over $$BD$$ such that on that chord $$BD$$ appears under $$\alpha$$. This in one locus for $$A$$. Pick a point $$A'$$ on the circle. Next step is to double $$A'D$$ over $$D$$ to get $$C'$$. If the choice of $$A'$$ meets the condition $$C'E=CE$$ then $$A'=A$$. It seems that the construction gives a unique triangle.