Need Help on this Problem

Zipsy · Apr 15, 2020

So there is a problem on the AHSME using mass points. It is the 1975 number 28 and there is no explanation on the art of problem solving website. This is the question:
In

$\triangle ABC$

shown in the adjoining figure,

$M$

is the midpoint of side

$BC, AB=12$

and

$AC=16$

. Points

$E$

and

$F$

are taken on

$AC$

and

$AB$

, respectively, and lines

$EF$

and

$AM$

intersect at

$G$

. If

$AE=2AF$

then

$\frac{EG}{GF}$

equals

$[asy] draw((0,0)--(12,0)--(14,7.75)--(0,0)); draw((0,0)--(13,3.875)); draw((5,0)--(8.75,4.84)); label(A, (0,0), S); label(B, (12,0), S); label(C, (14,7.75), E); label(E, (8.75,4.84), N); label(F, (5,0), S); label(M, (13,3.875), E); label(G, (7,1)); [/asy]$

If anyone can give me an explanation on how to do it, that would be great. I know the answer is 3/2 but I would like to know how to do it.

firemath · Apr 25, 2020

Are you familiar with the Angle Bisection Theorem?
(Click below if you aren't.....)

The "Angle Bisector" Theorem says that an angle bisector of a triangle will divide the opposite side into two segments that are proportional to the other two sides of the triangle. ... An angle bisector of a triangle divides the opposite side into two segments that are proportional to the other two sides of the triangle.

If you know that the answer is 3/2, then obviously EG=3 and GF=2 for your proportion.

Need Help on this Problem

Zipsy

New member

firemath

Full Member