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Geometry 14 Online
mathslover (mathslover):

P is any point on the median AD of ΔABC. Show that ar(APB) = ar(ACP)

mathslover (mathslover):

@ParthKohli

Parth (parthkohli):

ar?

mathslover (mathslover):

area

OpenStudy (amistre64):

congruent triangle thrms?

mathslover (mathslover):

yes sir

Parth (parthkohli):

APB and ACP are not necessarily congruent. :-\

mathslover (mathslover):

Well the chapter name was " Area of parallelograms and triangles " but I am sure that it will use congruent angle theorems or "mid-pt. theorem"

OpenStudy (amistre64):

|dw:1362761588096:dw|

mathslover (mathslover):

We know that : the median of a triangle divides a triangle into two congruent triangles ( equal areas ) ..

Parth (parthkohli):

A triangle \(b = 2 \) and \(h = 2\) has the same area of one with \(b = 4\) and \(h = 1\). So I can pretty much assure that they are not necessarily congruent

OpenStudy (amistre64):

the area of a triangle is height times base divided by 2 area of the left is hx/2, area of the right is hx/2

Parth (parthkohli):

LOL, nice! Didn't think of that one

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