Prove that the altitudes of a triangle are concurrent.?

Answer 1

Let's try a combination of geometry and calculus.

Let A,E,F,B be points on the circumference of a semicircle where AB = diameter. Because of the properties of circles, angles AFB and AEB are right angles.
So let C be the point of intersection of AE and BF (extended). A, B and C form the vertices of the triangle in question. BF, AE are two of the altitudes.
Let G be the point of intersection of AF and BE. So G is supposedly the point where the altitudes meet.
We are required to prove that CG is perpendicular to AB.

Without loss of generality, let A = (-1, 0) and B = (1,0) and C be any point in the upper halfplane. We need to show that C and G have the same x cordinate.

eqn of AE: y = (1+x) * √[(1-xE)/(1+xE)]
eqn of BF: y = (1-x) * √[(1+xF)/(1-xF)]
eqn of AF: y = (1+x) * √[(1-xF)/(1+xF)]
eqn of BE: y = (1-x) * √[(1+xE)/(1-xE)]

C is where AE and BF intersect, ie xC satisfies
(1+xC) * √[(1-xE)/(1+xE)] = (1-xC) * √[(1+xF)/(1-xF)]
or xC * {√[(1-xE)/(1+xE)] + √[(1+xF)/(1-xF)]}
= √[(1+xF)/(1-xF)] - √[(1-xE)/(1+xE)]

Multiply each term by √ (1+xE) * √ (1-xF)
xC * {√[(1-xE)*(1-xF)] + √[(1+xF)*(1+xE)]}
= √[(1+xF)*(1+xE)] - √[(1-xE)*(1-xF)]

Similarly G is where AF and BE intersect, ie xG satisfies the same equation as above with the xE and xF switched. And you'll notice that due to the symmetry of the equation, you'll end up getting the identical expression for xG.

So xG = xC, hence GC is perpendicular to AB and it is proved that all three altitudes pass through the same point.

Answer 2

Here's a proof that draws upon cyclic quadrilaterals and their properties.

Draw triangle ABC with altitudes BY and CZ. Let BY and CZ intersect at H. Draw a line connecting A to H, and drop an altitude from H to line BC at X (altitude HX). We wish to show that A, H, and X are collinear. Then AX would be an altitude of ABC that's concurrent with the other altitudes.

One way to show A, H, and X are collinear is to prove
At this point you'll probably want to take a second and digest the last couple steps in which we came up with "subgoals" that would lead us directly to a solution if we could prove them. Our most recent subgoal is to show that
In particular, look at cyclic quadrilaterals YZBC and AYHZ. From the former we get

(1)
From the latter we get


From (1) and (2) it follows that

Answer 3

you can denote the point of intersection of two altitudes as point A(there must be such an A), then link the other vertex (denote as B),and link B and A. the only thing you should do is prove that AB is a altitude.