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# Not sure why this is done this way (Pie Chart question) watch

1. (Original post by ByronicHero)
I was never in year 11.
take your poetic crap somewhere else Shakespeare
2. (Original post by Mayhem™)
take your poetic crap somewhere else Shakespeare
I mean it literally.
3. (Original post by ByronicHero)
I mean it literally.
elaborate
4. (Original post by Mayhem™)
elaborate
Edit: Why will nobody help me do a maths.
5. Je mange le foot
Wait this isn't the french thread...
*flies away*
6. (Original post by homeland.lsw)
Je mange le foot
Wait this isn't the french thread...
*flies away*

"voulez vous coucher avec moi ce soir ?"
7. (Original post by Mayhem™)

"voulez vous coucher avec moi ce soir ?"
PRSOM
8. (Original post by Mayhem™)

"voulez vous coucher avec moi ce soir ?"
ma prof de francais est une femme obèse donc je ne pense pas que je le dirais
9. (Original post by TeeEm)
Usually it would have been my pleasure
Zacken physicsmaths 16Characters....
please take over as I have been stuck on this PDE for the last 3 hours and I need to finish tonight
Nobody helped me

In an acute-angled triangle ABC the interior bisector of the angle A intersects BC at L and intersects the circumcircle of ABC again at N. From point L perpendiculars are drawn to AB and AC, the feet of these perpendiculars being K and M respectively. Could you please help me prove that the quadrilateral AKNM and the triangle ABC have equal areas?
10. (Original post by ByronicHero)
Nobody helped me

In an acute-angled triangle ABC the interior bisector of the angle A intersects BC at L and intersects the circumcircle of ABC again at N. From point L perpendiculars are drawn to AB and AC, the feet of these perpendiculars being K and M respectively. Could you please help me prove that the quadrilateral AKNM and the triangle ABC have equal areas?
what have you done so far?

Renzhi10122
11. (Original post by ByronicHero)
Nobody helped me

In an acute-angled triangle ABC the interior bisector of the angle A intersects BC at L and intersects the circumcircle of ABC again at N. From point L perpendiculars are drawn to AB and AC, the feet of these perpendiculars being K and M respectively. Could you please help me prove that the quadrilateral AKNM and the triangle ABC have equal areas?
I dnt even have internet so I cnt help. But note the arcs are of same legth and use simlar triangles. This a BMO1 question right?
12. (Original post by ByronicHero)
Nobody helped me

In an acute-angled triangle ABC the interior bisector of the angle A intersects BC at L and intersects the circumcircle of ABC again at N. From point L perpendiculars are drawn to AB and AC, the feet of these perpendiculars being K and M respectively. Could you please help me prove that the quadrilateral AKNM and the triangle ABC have equal areas?
Notice that this reduces down to proving that [KLB]+[MLC]=[KLN]+[MLN]. Now notice that the two pairs of triangles share a side, and KL=ML, and so you want to prove that KB+LC=the two perpendicular distances from N to KL and N to ML.

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