## BDMO NATIONAL 2014 : Secondary 7

Posts: 17
Joined: Thu Aug 29, 2013 6:33 pm
Location: Japan Garden City, Mohammadpur, Dhaka

### BDMO NATIONAL 2014 : Secondary 7

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 $\angle YCE = 22.5 , \angle PXQ = ?$

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.

*Mahi*
Posts: 1175
Joined: Wed Dec 29, 2010 12:46 pm
Location: 23.786228,90.354974
Contact:

### 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.)

Use $L^AT_EX$, It makes our work a lot easier!

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### 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 $Y \in CX$.

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BDMO NATIONAL 2014 : Secondary 7

The figure is given below which I used to solve the problem:
Screenshot_2018-02-23-21-03-51-1-1.png

samiul_samin
Posts: 1007
Joined: Sat Dec 09, 2017 1:32 pm

### Re: BDMO NATIONAL 2014 : Secondary 7

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 (15.3 KiB) Viewed 1473 times
Any one help me to complete the solution.