Page 1 of 1

Regional Olympiad 2016

Posted: Sat Dec 09, 2017 2:34 pm
by samiul_samin
I need help to solve this problem.

Re: Regional Olympiad 2016

Posted: Tue Feb 13, 2018 8:22 pm
by samiul_samin
I have already solved it.According to the question,$CE=ED$ .So,$2ED=CD$ and $2EF=BC$.The answer is $6$.

Re: Regional Olympiad 2016

Posted: Tue Feb 13, 2018 8:27 pm
by Tasnood
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, $EF=\frac{1}{2}BC=\frac{12}{2}=6$

Re: Regional Olympiad 2016

Posted: Tue Feb 19, 2019 11:23 am
by samiul_samin
samiul_samin wrote:
Sat Dec 09, 2017 2:34 pm
I need help to solve this problem.
Screenshot_2019-02-19-11-22-57-1.png
Screenshot_2019-02-19-11-22-57-1.png (81.04 KiB) Viewed 11535 times