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

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.

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

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.

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.

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