Barishal higher sec 10
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Mehfuj Zahir
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In triangle ABC angle BAC is 60.AD,BE,CF are the angle bisector of angle A,B.C respectively.Find the value of angle FEI&EFI?
Re: Barishal higher sec 10
wheres from 'I' comes in FEI and EFI?
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Mehfuj Zahir
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Re: Barishal higher sec 10
most probably it is their intersecting point.The question has not definition about it
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nafistiham
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Re: Barishal higher sec 10
as much as i understand i think the definition of i is not here because, we usually use i to be an in center of a triangle.when ad,be,cf intersect,it is the in center.may be that's why the definition is not here.
the answers are may be 30' and 30'
the answers are may be 30' and 30'
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Barishal higher sec 10
Tiham,
the solution is definitely $30^{\circ}$.
$I$ is meant to be the incentre here.
the solution is definitely $30^{\circ}$.
$I$ is meant to be the incentre here.
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Re: Barishal higher sec 10
Amlan da, Notice that $AEIF$ is a cyclic quadrangle. $(\angle EIF=120^{\circ})$
Then the result follows
Then the result follows
Last edited by Labib on Sat Dec 03, 2011 1:32 am, edited 1 time in total.
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Re: Barishal higher sec 10
$\angle BIF=120^{\circ}$ !!! or $\angle EIF=120^{\circ}$ ? how have u assumed it?Labib wrote: $AEIF$ is a cyclic quadrangle. $(\angle BIF=120^{\circ})$
Then the result follows
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Re: Barishal higher sec 10
Edited it.. Should be fine now.. 
$\frac 1 2 \angle B + \frac 1 2 \angle C=60^{\circ}$.
$\angle BIC=\angle EIF= 180^{\circ}-(\frac 12 \angle B + \frac 12 \angle C)=180^{\circ}-60^{\circ}=120^{\circ}$.
$\frac 1 2 \angle B + \frac 1 2 \angle C=60^{\circ}$.
$\angle BIC=\angle EIF= 180^{\circ}-(\frac 12 \angle B + \frac 12 \angle C)=180^{\circ}-60^{\circ}=120^{\circ}$.
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