This exercise outlines a proof of the Median Concurrence Theorem. Let ^ ABC be a

Question

This exercise outlines a proof of the Median Concurrence Theorem. Let ^ABC be a triangle, and let D, E, and F be the midpoints of the sides opposite A, B, and C, respectively. Let L = (line)AD and let m and n be the lines parallel to L that contain the points E and F, respectively. Let H be the point at which m crosses BC, let I be the point at which n crosses BC, and let J be the point at which n crosses BE.

(a)  Use the Crossbar Theorem to prove that any two of the medians must intersect.

(b)  Use the Parallel Projection Theorem to prove that CH = HD and DI = IB.

(c) Let G be the point at which the medians AD and BE intersect. Prove that

(d) Use the fact that part (c) is valid for the point of intersection of any two medians to conclude that all three medians intersect at G.

Explanation / Answer

1.

a)

consider the rays AB and BD , by the Cross bar theorem BE (is a ray from B) should intersect AD.

thus proved

2.

b)

we know that D,E,F are midpoints pf BC,AC,AB respectively

=>

BD = DC = BC/2, CE=EA = AC/2 , BF = FA = AB/2

L,m,n are parallel lines.

then by parallel projection theorem, we have

BI/BD = BF/BA = 1/2=> BI = BD/2, ID = BD-BI = BI

CH/CD = CE/CA =1/2 => CH =CD/2, HD = CD-CH = CH

the above two statemnts proves BI = ID=BD/2, CH =HD =CD/2 and we know that BD=DC =>

BI=ID=DH=HC

thus proved

c)

using the parallel projection theorem on L,n lines we have

BJ/BG = BF/BA = 1/2

=> BJ =BG/2, JG = BG-BJ = BJ

from above problem we know that BI=ID=DH=HC

=> BD =2BI, BH = 3BI

applying the parallel projection theorem on lines L, m, we have

BD/BH = BG/GE

=> 2BI/3BI = BG/GE

=> BG = 2/3.BE => 2BI = 2/3BE => BI = BE/3

GE = BE-BG = BE-2/3BE = BE/3 = BI

=> BI =IG = GE

thus proved

BG = BI+IG = GE+GE = 2GE

thus proved

d)

let medians AD, and BE meet at G, assume that the medians BE, CF meet at G1. we need to prove that G and G1 are same point

AD and BE meet at G => from above problem we have BG = 2/3BE,

CF and BE meet at G1 => from above problem we have BG1 = 2/3BE

from above two equations BG = BG1

=> G and G1 are same point .

=> medians AD<BE<CF meet at point G => all medians are concurrent.

thus proved.

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