Prove that the heights drawn from the vertices of the acute angles of an isosceles obtuse triangle are equal.

Let a triangle ABC be given, <B – obtuse, then the angles <A = <C – acute. Let’s draw the heights АН and СН1, we prove АН = СН1.

Triangles АHС and АСH1 are equal, on the basis of equality of sides and angles. The AC side is common to them, angles <АHС = <СH1А = 90 (since these are heights); angles <BAC = <BCA, as angles at the base of the AC. 2 angles are equal, hence the angles <CH1A = <AHC

Sides AH = CH1, as sides against equal angles <CH1A = <AHC.