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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.

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Q: Prove theorem the height of an isosceles triangle is the median?
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If a median of a triangle is also an altitude of a triangle then is the triangle isosceles?

Yes


Can the altitude from the vertex angle of an isosceles triangle be the median?

If the triangle is really isosceles, and it's not lying on one of the equal sides, then the altitude is always a median.


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