BDMO NATIONAL 2014 : Secondary 7
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In $ \triangle ABC$ $E, F$ are two points on $BC$ and $AB$ such that $ EF || AC . Q$ is a point on $AB$ such that $ \frac{AQ}{PQ} = \frac{30}{13} $ . $PQ$ is parallel to $EF$ where $P$ lies on $CB$ . $X$ is taken on extended $EQ$ such that $CX = 20.4$ . Given $ \frac{CY}{EY} = \frac{XY}{CY} , PX = 15.6$ ; if $\angle YCE = 22.5 , \angle PXQ = ?$
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- Posts:17
- Joined:Thu Aug 29, 2013 6:33 pm
- Location:Japan Garden City, Mohammadpur, Dhaka
Re: BDMO NATIONAL 2014 : Secondary 7
I posted this to see a solution . But still there isn't any . Is there any problem with this problem ?
Last edited by sadman sakib on Tue Mar 11, 2014 6:05 pm, edited 1 time in total.
Re: BDMO NATIONAL 2014 : Secondary 7
Actually, there is.
To solve this problem, you have to assume $Y \in EX$ (and there is another little trick, but that can be found out.)
To solve this problem, you have to assume $Y \in EX$ (and there is another little trick, but that can be found out.)
Please read Forum Guide and Rules before you post.
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
Use $L^AT_EX$, It makes our work a lot easier!
Nur Muhammad Shafiullah | Mahi
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Re: BDMO NATIONAL 2014 : Secondary 7
Starting of the solution
Please,any one help to get the full solution.I will try to add the figure of this problem where $Y \in CX$.
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Re: BDMO NATIONAL 2014 : Secondary 7
The figure is given below which I used to solve the problem:
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- Posts:1007
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Re: BDMO NATIONAL 2014 : Secondary 7
Any one help me to complete the solution.samiul_samin wrote: ↑Sat Feb 24, 2018 12:14 amThe figure is given below which I used to solve the problem:
Screenshot_2018-02-23-21-03-51-1-1.png