Regional Olympiad 2016

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samiul_samin
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Regional Olympiad 2016

Unread post by samiul_samin » Sat Dec 09, 2017 2:34 pm

I need help to solve this problem.
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samiul_samin
Posts: 276
Joined: Sat Dec 09, 2017 1:32 pm
Location: Shantinagar,Digharkanda,Mymensingh
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Re: Regional Olympiad 2016

Unread post by samiul_samin » Tue Feb 13, 2018 8:22 pm

I have already solved it.According to the question,$CE=ED$ .So,$2ED=CD$ and $2EF=BC$.The answer is $6$.
$\int^{\infty}_0 e^{-x} dx=1$

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Tasnood
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Re: Regional Olympiad 2016

Unread post by Tasnood » Tue Feb 13, 2018 8:27 pm

In $\triangle ABC,\angle BAC=50^\circ,\angle ACB=65^\circ$. So, $\angle ABC=180^\circ-50^\circ-65^\circ=65^\circ$
So, $\angle ABC=\angle ACB \Rightarrow AB=AC$
So, $AB=AC=AD$
Between $\triangle ABF$ and $\triangle ADF, AB=AD,AF=AF,\angle AFB=\angle AFD=90^\circ$ So, $\triangle ABF \cong \triangle ADF$
So,$BF=DF=\frac{1}{2}BD$

Between $\triangle AEC$ and $\triangle ADE, AC=AD, AE=AE, \angle AEC=\angle AED=90^\circ$ So, $\triangle AEC \cong \triangle ADE$
So, $CE=DE=\frac{1}{2}CD$

Between $\triangle DEF$ and $\triangle DCB, \angle FDE= \angle BDC, \frac {DF}{BD}=\frac {DE}{CD}=\frac{1}{2}$
So, $\triangle DEF \sim \triangle DCB$
Then, $EF=\frac{1}{2}BC=\frac{12}{2}=6$

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