SOLUTION: Prove lines containing the altitudes of any triangle are concurrent

Algebra ->  Algebra  -> Geometry-proofs -> SOLUTION: Prove lines containing the altitudes of any triangle are concurrent      Log On

 Ad: Algebrator™ solves your algebra problems and provides step-by-step explanations! Ad: Algebra Solved!™: algebra software solves algebra homework problems with step-by-step help!

 Geometry: Proofs in Geometry Solvers Lessons Answers archive Quiz In Depth

 Question 387008: Prove lines containing the altitudes of any triangle are concurrentAnswer by richard1234(5390)   (Show Source): You can put this solution on YOUR website!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, is not equal to 1 since the three segments are not concurrent. We can show that , , and using similarities within the right triangles ADC, AFC, etc. Multiplying all three equations, The right side of the equation cancels to 1, so 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.