Given two equal triangles ABC = A1B1C1, in which the angle A = angle A1, and the angles B and B1 are obtuse.

Given two equal triangles ABC = A1B1C1, in which the angle A = angle A1, and the angles B and B1 are obtuse. Prove that the distances from the vertices A and A1, respectively, to the straight lines BC and B1C1 are equal.

Let there be given two equal obtuse triangles ∆ ABC = ∆ A₁B₁C₁, for which ∠A = ∠A₁, and ∠В and ∠В₁ are obtuse. To prove that the distances from the vertices A and A₁, respectively, to the straight lines BC and B₁C₁ are equal, draw the heights AH and A₁H₁ and consider the resulting rectangular ∆ ABH and ∆ A₁B₁H₁.
In them AB = A₁B₁ (as equal corresponding sides in ∆ ABC = ∆ A₁B₁C₁).
∠АВН = ∠А₁B₁Н₁ (as adjacent to equal corresponding angles ∠ABC and ∠ A₁B₁C₁ in ∆ ABC = ∆ A₁B₁C₁).
Hence, ∆ ABН = ∆ A₁B₁Н₁ by 3 signs of equality of right-angled triangles (by hypotenuse and acute angle). Then the corresponding legs AH = A₁H₁. And this is the distance from the vertices A and A₁, respectively, to the straight lines BC and B₁C₁.
Q.E.D.