The Great Divide

Geometry Level 4

You have a triangle ABC and you draw a line from a point D on AC such that it is parallel to BC and it meets AB at E. Join EC and BD and these lines meet at F. AF meets BC at G.

If G C B C = a b \frac { GC }{ BC } =\frac { a }{ b } , find a + b a+b .


The answer is 3.

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Vishnu C by vishnu c , 22, India

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2 solutions

Vishnu C
May 11, 2015

Ceva's theorem is applicable here as AG, CD and BE are concurrent.

S o y o u h a v e A D D C = A E E C b y s i m i l a r t r i a n g l e l o g i c a n d C G × B E × A D = D C × G B × E A , u s i n g C e v a ′ s t h e o r e m . W e g e t , a f t e r c a n c e l l a t i o n , C G = G B . S o , a = 1 a n d b = 2 t o g i v e a + b = 3. So\quad you\quad have\quad \frac { AD }{ DC } =\frac { AE }{ EC } \quad by\quad similar\quad triangle\quad logic\\ and\quad CG\times BE\times AD=DC\times GB\times EA,\quad using\quad Ceva's\quad theorem.\\ \\ We\quad get,\quad after\quad cancellation,\quad CG=GB.\quad So,\\ a=1\quad and\quad b=2\quad to\quad give\quad a+b=\boxed { 3. } \

4 Helpful 0 Interesting 0 Brilliant 0 Confused
Li Yuelin
May 11, 2015

Assume that D and E are the midpoints of AB and AC respectively. Thus F is the centroid of ABC. It therefore follows that AG is a median, G is the midpoint of BC, and a/b = 1/2. 1+2 = 3

0 Helpful 0 Interesting 0 Brilliant 0 Confused

Can you prove it for a general DE? D or E need not be the midpoint.

vishnu c - 6 years, 1 month ago

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