If the altitude from one vertex of a triangle bisects the opposite side, prove that the triangle is isosceles.
Answer:
- We know that an altitude from a vertex to the opposite side is the perpendicular drawn from that vertex to the opposite side.
Let us now draw the altitude AD from the vertex A of △ABC to the opposite side BC. - We need to prove that the triangle is isosceles, i.e., AB=AC.
- In △ADB and △ADC, we have BD=DC [Given] AD=AD [Common] ∠ADB=∠ADC=90∘[As AD⊥BC]∴
- As the corresponding parts of congruent triangles are equal, we have AB = AC .
Thus, \triangle ABC is an isosceles triangle.