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 ^@\triangle ABC^@ to the opposite side ^@BC^@. - We need to prove that the triangle is isosceles, i.e., ^@AB = AC^@.
- In ^@ \triangle ADB ^@ and ^@ \triangle ADC ^@, we have @^ \begin{aligned} BD = DC && \text { [Given] } \\ AD = AD && \text{ [Common] } \\ \angle ADB = \angle ADC = 90^ \circ && \text{[As } AD \perp BC \text{]}\\ \therefore \triangle ADB \cong \triangle ADC && \text{ [By SAS congruence criterion] } \\ \end{aligned}@^
- As the corresponding parts of congruent triangles are equal, we have ^@ AB = AC ^@.
Thus, ^@ \triangle ABC ^@ is an isosceles triangle.
