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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.

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  • $\begingroup$ 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 '16 at 19:51
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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.

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  • $\begingroup$ how did you decided about point $C$? is it on a trigonometric circle? $\endgroup$ – bony Jul 24 '16 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 '16 at 12:37
  • $\begingroup$ And point $B$ is inside/outside that circle, right? $\endgroup$ – bony Jul 24 '16 at 12:38
  • $\begingroup$ @bony: That does not matter. $B,B'$ are points on $x$-axis. $\endgroup$ – mathlove Jul 24 '16 at 12:39
  • $\begingroup$ ok, i just want to draw the cases to myself. thanks! $\endgroup$ – bony Jul 24 '16 at 12:40
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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.

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