If the altitudes from the two vertices of a triangle to the opposite sides are equal, prove that the triangle is isosceles.
Answer:
- Let BD and CE be the altitudes from the vertices B and D of △ABC.
As the altitude of a triangle is perpendicular to the opposite side, we have BD⊥AC and CE⊥AB Also, we are told that the altitudes are of equal length. ⟹BD=CE - We need to prove that AB=AC.
- In △ADB and △AEC, we have BD=CE[Given]∠BAD=∠CAE[Common]∠ADB=∠AEC=90∘[BD⊥AC and CE⊥AB]∴ △ADB≅△AEC[By AAS criterion]
- As the corresponding parts of congruent triangles are equal, we have
AB=AC.
Hence, △ABC is isosceles.