Let $AD\cap (BFC) $ in points $P$ and $Q$ and let $AD\cap (ABE)=M$ then $MP=MQ$.

Question

Let $\triangle DEF$ be the medial triangle of $\triangle ABC$ with standaring notations. Let $AD\cap (BFC) $ in points $P$ and $Q$ and let $AD\cap (ABE)=M$ then $MP=MQ$.

Here is the diagram:

There was a non synthetic solution given here https://artofproblemsolving.com/community/c1213795h2168723p16462131

But I am interested in a synthetic solution.

Thanks in advance.

Answer 1

We have to prove $z=u+v$.

First remember that (median) $$AD^2= {2AB^2+2AC^2-BC^2\over 4}$$ so we have $$a^2+(t+z+u)^2 = 2c^2+2b^2\;\;\;\;(*)$$

• PoP of A on $C_2$:$\;\;\;2c^2=t(t+z+u+v)$
• PoP of D on $C_2$:$\;\;\;a^2=v(z+u)$
• PoP of C on $C_1$:$\;\;\;b^2=a(a+y)$
• PoP of D on $C_1$:$\;\;\;ay=u(t+z+u)$

We have to eliminate $y,a,b$ and $c$. From last two equation we get (we eliminate $y$) $$b^2=a^2+u(u+z+t)$$

Now pluging $c^2,b^2$ and $a^2$ in to $(*)$ we get:

$$(t+z+u)^2= t(t+z+u+v)+v(z+u)+2u(u+z+t)$$

After some simple manipulations we get $z=u+v$.