If the altitudes from the two vertices of a triangle to the opposite sides are equal, prove that the triangle is isosceles.


Answer:


Step by Step Explanation:
  1. Let ^@BD^@ and ^@CE^@ be the altitudes from the vertices ^@B^@ and ^@D^@ of ^@\triangle ABC.^@

    As the altitude of a triangle is perpendicular to the opposite side, we have @^BD \perp AC \text { and } CE \perp AB @^ Also, we are told that the altitudes are of equal length. @^ \implies BD = CE @^
      A B C E D
  2. We need to prove that ^@AB = AC.^@
  3. In ^@\triangle ADB^@ and ^@\triangle AEC^@, we have @^ \begin{aligned} &BD = CE &&[\text{Given}] \\ &\angle BAD = \angle CAE &&[\text{Common}]\\ &\angle ADB = \angle AEC = 90^ \circ &&[ BD \perp AC \text { and } CE \perp AB] \\ \therefore \space &\triangle ADB \cong \triangle AEC &&[\text{By AAS criterion}] \end{aligned}@^
  4. As the corresponding parts of congruent triangles are equal, we have ^@AB = AC .^@
    Hence, ^@\triangle ABC^@ is isosceles.

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