Problem:
In the adjoining figure, the circle meets the sides of an equilateral triangle at six points. If AG=2,GF=13,FC=1 and HJ=7, then DE equals
Answer Choices:
A. 222​
B. 73​
C. 9
D. 10
E. 13
Solution:
In the adjoining figure, let AH=y,BD=a,DE=x and EC=b. We are given AG=2,GF=13,HJ=7 and FC=1. Thus the length of the side of the equilateral triangle is 16. Also, using the theorem on secants drawn to a circle from an external point, we have y(y+7)=2(2+13), or 0=y2+7y−30=(y−3)(y+10). Hence y=3 and BJ=6. Using the same theorem we have b(b+x)=1(1+13)=14 and a(a+x)=6(6+7)=78. Also, a+b+x=16. Solving these last three equations simultaneously gives x=222​ (also b=6−22​, a=10−22​).
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