Question 387008
Suppose triangle ABC has altitudes AD, BE, CF. Without loss of generality, let AD and BE meet at H. We want to prove that H lies on CF.


On the contrary, suppose H is not on CF. Then, by Ceva's theorem,


{{{(AF/FB)(BD/DC)(CE/CA)}}} is not equal to 1 since the three segments are not concurrent.


We can show that {{{CE/CD = BE/AD}}}, {{{AF/AE = FC/BE}}}, and {{{BD/BF = AD/FC}}} using similarities within the right triangles ADC, AFC, etc. Multiplying all three equations,


{{{(CE/CD)(AF/AE)(BD/BF) = (BE/AD)(FC/BE)(AD/FC)}}} The right side of the equation cancels to 1, so


{{{(CE/CD)(AF/AE)(BD/BF) = 1}}}


However, the left side of the equation is equivalent to our first equation, so 1 is equal to some quantity other than 1, contradiction. Therefore the three altitudes must be concurrent.