In the following figure, prove that $AC$ bisects $GH$.

Question

It is given to us that $ABCD$ is a parallelogram. $E$ and $F$ are mid- points of $CD$ and $BC$ respectively. $AC$ and $BD$ are it's diagonals. We have to prove that $AC$ bisects $GH$.

One may start by observing that $AC$ and $GH$ are diagonals of the quadrilateral $AHCG$. So proving it a parallelogram solves the problem. However, when I tried to prove it by using congruence, I ended up with one equality less. Similarity didn't work either. Can I get some hints for solving this?

My attempt:

I tried to prove that $AHCG$ is a parallelogram by proving $\Delta IGC \cong \Delta IHA$. The relations are $$IA = IC$$ $$\angle AIH = \angle CIG$$

However, there is no third relation. If I somehow prove that $GC = HA$ i can prove that $AHCG$ is a parallelogram. But I can't seem to get any new relation.

P.S. - I don't have a ton of knowledge in mathematics. An answer using elementary math would be appreciated.

Answer 1

Apply an affine transformation to map ABCD into a square. Affine transformation preserve midpoints of line segments, and the situation is now symmetric across the AC diagonal.

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Alternatively: $G$ is the point where the medians in triangle $ABC$ meet, and we know that the medians in any triangle cut each other in the ratio $2:1$. So $GI=\frac13BI$ and $HI=\frac13DI$, but $BI=DI$ because $ABCD$ is a parallellogram.

Answer 2

Fact #1 Diagonals of a parallelogram bisect each other.

Fact #2 The centroid of a triangle divides each median in the ratio 2 : 1.

G is the centroid of $\triangle ABC$. Then, GI = $\dfrac {1}{3} BI$

This is the same for H. That is $HI = \dfrac {1}{3} DI$

Result follows because of Fact #1.