Proving two triangles congruent given two congruent sides and a congruent median

Question

The title was a bit too short for me to fit the full details, so here's the scenario I have. Prove that two triangles are congruent if in two triangles, the median from the common vertex and two sides is congruent with the corresponding median and the two sides in the second triangle.

Trying to figure this question out for my geometry class but unsure about exactly what I still need. We learned about SAS, ASA, and SSS in our last unit. The wording I'm a bit confused on, is it saying that all 4 sides are congruent or just that the corresponding side on each triangle is congruent with the other corresponding side? I can tell we'll have to use either SAS or SSS for this question as we know 2 of the sides are congruent on each triangle but I'm not fully sure of which congruency method I need to use yet. I'll include the most recent theorem we learned in our last lesson. Any help on how to solve this would be appreciated!

1. The median to the base of a triangle is also the angle bisector to the angle at the vertex opposite the base of this triangle in an isosceles triangle.

Answer 1

Suppose you extend the median $AD$ in triangle $ABC$ (with sides $a$, $b$, $c$) beyond the point of intersection $D$ with the side $BC$ by an amount, equal to the size of the median $m$, giving the segment $AE$. Join $E$ with $B$ and $C$. It is easy to see that $ABEC$ is a parallelogram. His diagonals are $d_1=AE=2m$ and $d_2=a$. By the parallelogram identity, $$ 2(b^2+c^2)=d_1^2+d_2^2=(2m)^2+a^2 $$ Therefore, sides $b$, $c$ and $d_1=2m$ uniquely determine side $a$. As a consequence, your data completely determine the triangle.

Answer 2

First some clarification on the problem. It is not required that the four sides are all congruent. Let $\triangle ABC$ with median $AM$ and $\triangle A'B'C'$ with median $A'M'$ be the triangles in hypothesis. Then we know that $AB \cong A'B'$, $AC \cong A'C'$, and $AM \cong A'M'$.

(Similar to the other answer, but probably more closely related to the material you mention.)

Produce $AM$ to $D$ so that $AM \cong MD$ and $A'M'$ to $D'$ so that $A'M'\cong M'D'$.

By SAS criterion you have the following congruences: $\triangle AMD \cong \triangle CMB$, $\triangle ACM \cong \triangle DBM$, and analogously on the other triangle, $\triangle A'M'D' \cong \triangle C'M'B'$, $\triangle A'C'M' \cong \triangle D'B'M'$.

Given the hypotheses, and what we just said, by SSS criterion we have, e.g., $\triangle ACD \cong \triangle A'C'D'$.

Then the median of the above mentioned triangles must be congruent, that is $AM \cong A'M'$. Hence $AB \cong A'B'$, and finally the thesis, by SSS criterion.