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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
$\dfrac{AQ}{FQ}=\dfrac {30}{13}$
$\Rightarrow \dfrac {AF}{FQ}+1=\dfrac {30}{13}$
$\Rightarrow \dfrac {AF}{FQ}=\dfrac {17}{13}$
$\Rightarrow \dfrac {AF}{FQ}=\dfrac {20.4}{15.6}$
$\Rightarrow \dfrac {AF}{FQ}=\dfrac {CX}{PX}$... ... ... (1)

AS, $AC\parallel EF$, $\dfrac {AF}{FB} =\dfrac {CE}{EB}$... ... ... (2)

And $EF\parallel QP$,$\dfrac {FQ}{FB}=\dfrac {EP}{EB}$... ... ...(3)

(2)×(3), $\dfrac {AF}{FQ}=\dfrac {CE}{EP}$... ... ...(4)

From (1) and (4), $\dfrac {CE}{EP}=\dfrac {CX}{PX}$... ... ...(5)

It is given that $\dfrac {CY}{EY}=\dfrac {XY}{CY}$... ... ...(6)

How to interupt (5) & (6) to get the value of $\angle PXQ$?
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
Screenshot_2019-02-19-17-36-49-1.png (15.3 KiB) Viewed 1614 times
Any one help me to complete the solution.