Prove theorem the height of an isosceles triangle is the median?

Answer 1

The theorem is only true if the base is the side of different length.

To see this consider the right angled isosceles triangle with sides 1, 1 and √2. If one of the sides of length 1 is the base, the height is obviously the other side of length 1, but it clearly does not meet the base at its mid-point to make it a median.

So with an isosceles triangle ABC with sides AB & AC equal, angles ABC & ACB equal and side BC the base, we need to prove that the point X where the height (AX) meets BC is such that BX = CX.

Consider triangles AXB and AXC.

1. Angle AXB is a right angle, as is AXC (since AX is a height of triangle ABC).
2. Side AB is the hypotenuse of triangle AXB; AC is the hypotenuse of triangle AXC; they are known to be equal (from isosceles triangle ABC)
3. Side AX is common to both triangles

Thus triangles AXB and AXC are congruent since we have a Right-angle, Hypotenuse, Side match.

Thus XB must be the same length as XC, that is X is the mid-point of BC.

As X is the mid-point of BC, AX is the median.