Angle Bisector and Altitude

Geometry Level 3

If A B = 13 AB = 13 , B C = 14 BC = 14 , and A C = 15 AC = 15 , D F DF can be written in the form a b c \frac{a\sqrt{b}}{c} , where a a and c c are coprime, positive integers and b b is square-free. What is a + b + c a+b+c ?


The answer is 35.

Aidan Poor by Aidan Poor , 18, USA

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2 solutions

Aidan Poor
Oct 7, 2018

Hint: Angle Bisector Theorem and similar triangles.

2 Helpful 1 Interesting 0 Brilliant 1 Confused

Please post the complete solution

Bivab Mishra - 2 years, 8 months ago
Ajit Athle
Oct 9, 2018

We can begin with the fact that, BE=5 & EC=9 and BD²= 14*13(1- 15 ² 27 ² \frac{15²}{27²} ) or BD= 28 √ 13 9 \frac{28√13}{9} . Also, C D D A \frac{CD}{DA} = 14 13 \frac{14}{13} or C A A D \frac{CA}{AD} = 27 13 \frac{27}{13} . Using Menelaus's Theorem in Tr. BDC with AFD as the transversal, [ B E E C \frac{BE}{EC} ][ C A A D \frac{CA}{AD} ][ D L L B \frac{DL}{LB} ]=1 which yields, D L L B \frac{DL}{LB} = 13 15 \frac{13}{15} or D L B D \frac{DL}{BD} = 13 28 \frac{13}{28} . Then DL = 13 √ 13 9 \frac{13√13}{9}

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