Prove that the perpendiculars drawn from the vertices of equal a | Math Question and Answer

Answer:

Step by Step Explanation:

Let △ABC be an isosceles triangle with ∠B = ∠C. Now, let us draw the perpendiculars from ∠B and ∠C to the opposite sides.

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

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