Problem:
In β³ABC,D is on AC and F is on BC. Also, ABβ₯AC,AFβ₯BC, and BD=DC=FC=1. Find AC.
Answer Choices:
A. 2β
B. 3β
C. 32β
D. 33β
E. 43β
Solution:
Let AC=x and β DCF=ΞΈ. Then β CBD=ΞΈ, and β ADB=2ΞΈ by the Exterior Angle Theorem. Thus cosΞΈ=x1β and cos2ΞΈ=xβ1. Therefore,
2(x1β)2β12βx2xβ=xβ1=x3βx2=32ββ
OR
Drop DGβ₯BC. Let AC=x,GC=y. Note that BC=2y, for β³BDC is isosceles. Since β³DCGβΌβ³ACFβΌβ³BCA, we obtain y1β=1xβ=x2yβ. Thus y=x1β and y=2x2β, implying x3=2, or x=32β.
.jpg)
.jpg)