Problem:
In the adjoining figure triangle ABC is inscribed in a circle. Point D lies on AC with DC=30∘, and point G lies on BA with BG>GA. Side AB and side AC each has length equal to the length of chord DG and ∠CAB=30∘. Chord DG intersects sides AC and AB at E and F, respectively. The ratio of the area of △AFE to the area of △ABC is
Answer Choices:
A. 32−3
B. 323−3
C. 73−12
D. 33−5
E. 39−53
Solution:
In the adjoining figure, line segment DC is drawn. Since AC=150∘,AD=AC−DC=150∘−30∘=120∘. Hence ∠ACD=60∘. Since AC=DG, GA=GD−AD=AC−120∘=30∘. Therefore, CG=180∘ and △CDG=90∘. Thus △DEC is a 30∘−60∘−90∘ right triangle.
Since we are looking for the ratio of the areas, let us assume without loss of generality that AC=AB=DG=1.
Let AE=x=DE. Then CE=1−x=3x⋅2. Solving for x yields AE=x=23−3. Let FH be the altitude of △AFE on AE. Now EH=2AE=223−3 and FH=(223−3)33.
The area of △AFE=(EH)(FH)=(223−3)233=473−12. Also, area △ABC=21(AB)(AC)sin30∘=21⋅(1)⋅(1)⋅21=41. Hence
