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BDMO NATIONAL 2014 : Secondary 7
Posted: Wed Feb 26, 2014 6:29 pm
by sadman sakib
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 $\angle YCE = 22.5 , \angle PXQ = ?$
Re: BDMO NATIONAL 2014 : Secondary 7
Posted: Sun Mar 09, 2014 9:15 pm
by sadman sakib
I posted this to see a solution . But still there isn't any . Is there any problem with this problem ?
Re: BDMO NATIONAL 2014 : Secondary 7
Posted: Sun Mar 09, 2014 11:24 pm
by *Mahi*
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.)
Re: BDMO NATIONAL 2014 : Secondary 7
Posted: Tue Feb 20, 2018 11:01 am
by samiul_samin
Starting of the solution
Please,any one help to get the full solution.I will try to add the figure of this problem where $Y \in CX$.
Re: BDMO NATIONAL 2014 : Secondary 7
Posted: Sat Feb 24, 2018 12:14 am
by samiul_samin
The figure is given below which I used to solve the problem:
Screenshot_2018-02-23-21-03-51-1-1.png
Re: BDMO NATIONAL 2014 : Secondary 7
Posted: Tue Feb 19, 2019 5:39 pm
by samiul_samin
samiul_samin wrote: ↑Sat Feb 24, 2018 12:14 am
The figure is given below which I used to solve the problem:
Screenshot_2018-02-23-21-03-51-1-1.png
Screenshot_2019-02-19-17-36-49-1.png
Any one help me to complete the solution.