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This is only true of triangles. Rhombi and other "squashed" polygons with more than three sides show that it is not true otherwise.

The two equal sides meet at an angle. It can be shown that the bisector of that angle divides the triangle into two triangles with one set of equal sides, one common side and these sides define angles of equal measure. So by SAS, the two triangle are congruent and so the angles in question are equal.

Alternatively, you could prove (as easily) that the altitude from that angle divides the original triangle into two right angled triangles with a common side and equal hypotenuses. Again congruence resulting in the equality of the angles as required.

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14y ago

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Q: Why are angles equal when sides are equal?
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