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samiul_samin
Posts: 276
Joined: Sat Dec 09, 2017 1:32 pm
Location: Shantinagar,Digharkanda,Mymensingh
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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
Contact:

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

Tasnood
Posts: 46
Joined: Tue Jan 06, 2015 1:46 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, $EF=\frac{1}{2}BC=\frac{12}{2}=6$