proving that triangles $ABC$, $A'B'C'$ are congruent

Question

Given $AD$ is a median to $BC$ in triangle $ABC$, and $A'D'$ is a median to $B'C'$ in triangle $A'B'C'$, and $AD=A'D', AC=A'C', AB=A'B'$.

How can i prove that triangles $ABC$, $A'B'C'$ are congruent?

I can't see how the median is helping me to prove that.

I tried to build a Parallelogram but it didn't work out.

Thanks.

Answer 1

From $AB =A'B’$, we can let AB and A’B’ be the same line. Other lines meeting the requirement are drawn as shown.

As mentioned, we form the parallelograms CABX and C’A'B'X’.

By SSS, $\triangle ABX \cong \triangle A’B’X’$. Then, the green marked angles are equal. In turn, the red marked angles are also equal.

Result follows by applying SAS.

Answer 2

We may suppose that $$A(0,0),\quad B(c,0),\quad C(b\cos\alpha,b\sin\alpha)$$ $$A'(0,0),\quad B'(c,0),\quad C'(b\cos\beta,b\sin\beta)$$ where $b\gt 0,c\gt 0,0\lt\alpha\lt \pi,0\lt\beta\lt\pi$. Then, $$D\left(\frac{c+b\cos\alpha}{2},\frac{b\sin\alpha}{2}\right),\quad D'\left(\frac{c+b\cos\beta}{2},\frac{b\sin\beta}{2}\right)$$

Now $$\begin{align}AD=AD'&\implies \left(\frac{c+b\cos\alpha}{2}\right)^2+\left(\frac{b\sin\alpha}{2}\right)^2=\left(\frac{c+b\cos\beta}{2}\right)^2+\left(\frac{b\sin\beta}{2}\right)^2\\&\implies c^2+2bc\cos\alpha+b^2=c^2+2bc\cos\beta+b^2\\&\implies \cos\alpha=\cos\beta\\&\implies \alpha=\beta\end{align}$$ from which $$\angle{BAC}=\angle{B'A'C'}$$ follows.

Answer 3

An answer by pure euclidian geometry:

Extend $AD$ to $E$ that $AD=ED$, extend $A'D'$ to $E'$ that $A'D'=E'D'$.
$AD$ is a median to $BC$ and $A'D'$ is a median to $BC$, so $BD=CD$ and $B'D'=C'D'$.
Also, $\angle ADB=\angle EDC$ and $\angle A'D'B'=\angle E'D'C'$.
By SAS, $\triangle ABD\cong\triangle ECD$ and $\triangle A'B'D'\cong\triangle E'C'D'$.
So $CE=AB$ and $C'E'=A'B'$, and by SSS, $\triangle ACE\cong\triangle A'C'E'$.
So $\angle ACE=\angle A'C'E'$, that is, $\angle ACB+\angle ECD=\angle A'C'B'=\angle E'C'D'$.
$\triangle ABD\cong\triangle ECD$ and $\triangle A'B'D'\cong\triangle E'C'D'$, so $\angle ABD=\angle ECD$ and $\angle A'B'D'=\angle E'C'D'$.
So $\angle ACB+\angle ABD=\angle A'C'B'=\angle A'B'D'$.
So $\angle BAC=\angle B'A'C'$, and by SAS, $\triangle ABC\cong\triangle A'B'C'$.

Note: if there is a median in the question, try double it to construct a pair of congruent triangles.