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.


  • $\begingroup$ Hint: Extend $AD$ to say $AD'$ such that $AD=DD'$, so that $ABD'C$ becomes a parallelogram. Then you can search for congruence. $\endgroup$
    – Sawarnik
    Jul 24, 2016 at 19:51

3 Answers 3


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

enter image description here

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.


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.

  • $\begingroup$ how did you decided about point $C$? is it on a trigonometric circle? $\endgroup$
    – bony
    Jul 24, 2016 at 12:35
  • $\begingroup$ @bony : $C,C'$ are points on the circle whose center is $(0,0)$ with radius $b$. They satisfy $x^2+y^2=b^2$. $\endgroup$
    – mathlove
    Jul 24, 2016 at 12:37
  • $\begingroup$ And point $B$ is inside/outside that circle, right? $\endgroup$
    – bony
    Jul 24, 2016 at 12:38
  • $\begingroup$ @bony: That does not matter. $B,B'$ are points on $x$-axis. $\endgroup$
    – mathlove
    Jul 24, 2016 at 12:39
  • $\begingroup$ ok, i just want to draw the cases to myself. thanks! $\endgroup$
    – bony
    Jul 24, 2016 at 12:40

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.


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