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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 (15.3 KiB) Viewed 1614 times

Any one help me to complete the solution.