A geometry problem by Edgar de Asis Jr.
ABC and DBC are right triangles. AC =11 , DB = 4 and BC = 8. AB and DC intersect at E, and F is the foot of the perpendicular from E to BC.
If EF = a/b, where a, b are positive coprime integers, what is a+b?
The answer is 59.
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3 solutions
E F 1 = A C 1 + B D 1 (note: This is a derived equation)
E F 1 = 1 1 1 + 4 1
E F 1 = 4 4 1 5
Thus,
E F = 1 5 4 4
The desired answer is 1 5 + 4 4 = 5 9
I got it right though it should be a+b
i stand to be corrected. thanks anyway
It's the Thales' theorem and we have two relationship:
EF/AC=BF/BC
EF/DB=CF/BC
we add this two relationships together:
EF/AC+EF/DB=BF/BC+CF/BC EF/11+EF/4=BC/BC=1
15EF/44=1
EF=44/15=a/b
a+b=59