Prove that the perpendiculars drawn from the vertices of equal angles of an isosceles triangle to the opposite sides are equal.


Answer:


Step by Step Explanation:
  1. Let ^@\triangle ABC^@ be an isosceles triangle with ^@\angle B^@ = ^@\angle C^@. Now, let us draw the perpendiculars from ^@ \angle B^@ and ^@ \angle C^@ to the opposite sides.

    Thus, ^@ BD \perp AC ^@ and ^@ CE \perp AB. ^@
      A B C E D
  2. We need to prove that ^@BD = CE.^@
  3. In ^@\triangle BCD^@ and ^@\triangle BCE,^@ we have @^ \begin{aligned} & BC = BC && [\text{Common}] \\ & \angle BDC = \angle CEB && [\text{Each 90} ^ \circ] \\ & \angle BCD = \angle CBE && [ \text{As} \space ABC \space \text{is an isosceles triangle.]} \\ & \therefore \space \triangle BCD \cong \triangle BCE && [\text{By AAS criterion}] \end{aligned} @^
  4. As the corresponding parts of congruent triangles are equal, we have @^BD = CE @^
  5. Thus, the perpendiculars drawn from the vertices of equal angles of an isosceles triangle to the opposite sides are equal.

You can reuse this answer
Creative Commons License

What they say