Problem: 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 EG/GF equals
Answer Choices:
A. 3/2
B. 4/3
C. 5/4
D. 6/5
E. not enough information given to solve the problem
Solution:
Construct line CP parallel to EF and intersecting AB at P. By proportionality AP =8. Let a,x,y,Ξ±,Ξ²,Ξ΄ and ΞΈ be as shown in the adjoining diagram. Then
sinΞ±aβ=sinΞΈ12β;sinΞ²aβ=sin(180ββΞΈ)16β;sinΞ±sinΞ²β=43β.
Therefore,
sinΞ±xβ=sinΞ΄8β;sinΞ²yβ=sin(180ββΞ΄)16β;
and
GFEGβ=xyβ=2sinΞ±sinΞ²β=23β.β
OR
In the adjoining figure side BC extended through C intersects EF extended through E at H. Applying Menelaus' theorem to β³FBH and β³HCE with AM as the transversal, we obtain
GFGHβABAFβMHMBβ=1,GEGHβACAEβMHMCβ=1,
respectively. Since MC=MB and AE=2AF, dividing the first equation by the second yields
GFGEβ=2ACABβ=2β
1612β=23β.
Menelaus' theorem: The six segments determined by a transversal on the sides of a triangle are such that the product of three non-consecutive segments is equal to the product of the remaining three.
.jpg)
.jpg)
