Prove that lines passing through the midpoints of sides of a triangle and the midpoints of cevians are also concurrent

Question

$AD, BE$ and $CF$ are three concurrent lines in $\triangle ABC$, meeting the opposite sides in $D, E$ and $F$ respectively. Show that the joins of the midpoints of $BC, CA$ and $AB$ to the midpoints of $AD, BE$ and $CF$ are concurrent.

Let $D', E'$ and $F'$ be the midpoints of $BC, CA$ and $AB$. Then consider $\triangle ABD$. $E'F' \parallel BC$ and therefore in $\triangle ABD$, $E'F'$ cuts $AD$ at its midpoint (By the midpoint theorem). Similiarly $E'D'$ cuts $CF$ at its midpoint and $D'F'$ cuts $BC$ at its midpoint.

How do I proceed with this?

Answer 1

Let us assume that $AL_A, BL_B, CL_C$ are three cevians in $ABC$, $M_A,M_B,M_C$ are the midpoints of the sides of $ABC$ and $N_A,N_B,N_C$ are the midpoints of the cevians from $A,B,C$.

Of course $N_A,N_B,N_C$ lie on the sides of the medial triangle $M_A M_B M_C$. $ABC$ and $M_A M_B M_C$ are perspective by the existence of the centroid. $N_A N_B N_C$ and $ABC$ are perspective since $AL_A,BL_B,CL_C$ are cevians. By transitivity $N_A N_B N_C$ and $M_A M_B M_C$ are perspective, hence the lines $M_A N_A, M_B N_N, M_C N_C$ are concurrent.

TLDR: the perspectivity is transitive.

Answer 2

Let $P$, $Q$ and $R$ be the midpoints of $AD$, $BE$ and $CF$ respectively. As OP pointed out, $P$ lies on $E'F'$, $Q$ lies on $F'D'$ and $R$ lies on $D'E'$.

By Ceva's Theorem,

$$\frac{BF}{FA}\cdot\frac{AE}{EC}\cdot\frac{CD}{DB}=1$$

By the midpoint theorem, $\displaystyle D'R=\frac{1}{2}BF$, $\displaystyle RE'=\frac{1}{2}FA$, $\displaystyle E'P=\frac{1}{2}CD$, $\displaystyle PF'=\frac{1}{2}BD$, $\displaystyle F'Q=\frac{1}{2}AE$ and $\displaystyle QD'=\frac{1}{2}EC$.

So, we have

$$\frac{D'R}{RE'}\cdot\frac{E'P}{PF'}\cdot\frac{F'Q}{QD'}=1$$

By the converse of Ceva's Theorem, $D'P$, $E'Q$ and $F'R$ are concurrent.