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 $BE=CD$ .
Easy-quizy
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Re: Easy-quizy
Well, there are many approaches. I tried Appolonius' Theorem. Similarity also gives a nice solution.
- nafistiham
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Re: Easy-quizy
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
for the reverse case, I have used a different way, see if it works
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Easy-quizy
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 $\triangle BGD \cong \triangle CGE \Rightarrow BD=CE \Rightarrow AB=AC$.
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 $\triangle BGD \cong \triangle CGE \Rightarrow BD=CE \Rightarrow AB=AC$.
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Re: Easy-quizy
I also used Apollopnius for the reverse proof.But these proofs are smarter!sowmitra wrote:Well, there are many approaches. I tried Appolonius' Theorem. Similarity also gives a nice solution.
An amount of certain opposition is a great help to a man.Kites rise against,not with,the wind.
Re: Easy-quizy
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 $BD=EC \rightarrow 2BD=2EC \rightarrow AB=AC$
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 $BD=EC \rightarrow 2BD=2EC \rightarrow AB=AC$
Re: Easy-quizy
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.sowmitra wrote: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 $BD=EC \rightarrow 2BD=2EC \rightarrow AB=AC$
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