Prove that if the lengths of two altitudes of a triangle are equal, then the triangle is isosceles.

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Solution

Let the altitudes $h_{a}$ and $h_{b}$ of the triangle $△ A B C$ be equal.

If R is the Circumradius of the triangle, then we know that the altitudes are given by the formula $h_{a} = \frac{b c}{2 R}$ and $h_{b} = \frac{a c}{2 R}$

$h_{a} = h_{b} ⟹ \frac{b c}{2 R} = \frac{a c}{2 R}$
$⟹ b = a$

Thus the given triangle must be isosceles.

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