Easy-quizy

[sakibtanvir]

That is easy but not a spoiler!,I think.
Problem: In $\Delta ABC$, $D$ and $E$ are the midpoints of $AB$ and $AC$ respectively.Prove that, $AB=AC$ if and only if $BE=CD$ .

[sowmitra]

Well, there are many approaches. I tried Appolonius' Theorem. Similarity also gives a nice solution.

[nafistiham]

Its is easy to prove that, when $AB=AC$, $BE=CD$ by showing $\triangle ABE \cong \triangle ACD$
for the reverse case, I have used a different way, see if it works

Suppose, $BE=CD$ and $AB$ is inequal to (\neq is not working ) $AC$ then let, $AB>AC$
Let us take the reflection of point $A$ across the perpendicular bisector of $BC$, which is ${A}'$
Add ${A}',B ; {A}',C$ mid points of ${A}'B,{A}'C$ are ${D}',{E}'$
Add ${E}',B ; {D}',C$
$\triangle ABC \cong \triangle {A}'CB$
meaning, $BE=B{E}'=CD=C{D}'$
But, $E,D,{E}',{D}'$ are at the same height
so, $BE$ is inequal to $B{E}'$
contradiction

Well, I there is a lot of handwork here. But, just if you figure out the figure in your brain, then you don't need even similarity !!

[Tahmid Hasan]

Proof of the converse
Let $BE \cap CD=G$
Now $BE=CD \Rightarrow \frac{2}{3}BE=\frac{2}{3}CD \Rightarrow BG=CG$
$BE=CD \Rightarrow \frac{1}{3}BE=\frac{1}{3}CD \Rightarrow GE=GD$
Hence $\triangle BGD \cong \triangle CGE \Rightarrow BD=CE \Rightarrow AB=AC$.

[sakibtanvir]

Well, there are many approaches. I tried Appolonius' Theorem. Similarity also gives a nice solution.

I also used Apollopnius for the reverse proof.But these proofs are smarter!

[sowmitra]

I have got another synthetic proof:
Drop perpendiculars $DM$ and $EN$ on to $BC$ from $D$ and $E$.
Now, $DM=EN$ because they are between the same parallel lines $DE$ and $BC$.
In the right-angled $\triangle^s DMC,ENB$: $DC=BE$ and $DM=EN$.
So, $\triangle DMC \cong \triangle ENB $
$\rightarrow MC=NB \rightarrow BC-MC=BC-NB \rightarrow BM=NC$
This again implies that $\triangle DBM \cong \triangle ENC,$
which implies $BD=EC \rightarrow 2BD=2EC \rightarrow AB=AC$

[SANZEED]

I have got another synthetic proof:
Drop perpendiculars $DM$ and $EN$ on to $BC$ from $D$ and $E$.
Now, $DM=EN$ because they are between the same parallel lines $DE$ and $BC$.
In the right-angled $\triangle^s DMC,ENB$: $DC=BE$ and $DM=EN$.
So, $\triangle DMC \cong \triangle ENB $
$\rightarrow MC=NB \rightarrow BC-MC=BC-NB \rightarrow BM=NC$
This again implies that $\triangle DBM \cong \triangle ENC,$
which implies $BD=EC \rightarrow 2BD=2EC \rightarrow AB=AC$

Well,after proving that $\triangle DMC \cong \triangle ENB $,we can also use the fact that $\angle DCM=\angle EBN$. This along with $BE=CD$ and $BC$ being common implies that $\triangle BCD\cong \triangle CBE$. Then $BD=CE$ and the result follows.