If the altitude from one vertex of a triangle bisects the opposite side, prove that the triangle is isosceles.


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Answer:


Step by Step Explanation:
  1. 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^@.
      A B C D
    As the altitude ^@AD^@ bisects the opposite side ^@BC^@, @^BD = DC@^.
  2. We need to prove that the triangle is isosceles, i.e., ^@AB = AC^@.
  3. 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}@^
  4. As the corresponding parts of congruent triangles are equal, we have ^@ AB = AC ^@.
    Thus, ^@ \triangle ABC ^@ is an isosceles triangle.

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