Question

if from any point on the base of an isosceles triangle perpendicular are drawn to the equal sides prove that the sum of these perpendiculars is equal to the altitudes on either of the equal angles​

Answer 1

Answer:

Let ABC be an isosceles triangle such that AB=AC.

Let AD be the bisector of ∠A.

To prove:- BD=DC

Proof:-

In △ABD&△ACD

AB=AC(∵△ABC is an isosceles triangle)

∠BAD=∠CAD(∵AD is the bisector of ∠A)

AD=AD(Common)

By S.A.S.-

△ABD≅△ACD

By corresponding parts of congruent triangles-

⇒BD=DC

Hence proved that the perpendicular drawn from the vertex angle to the base bisect the vertex angle and base.

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Answer 2

Hence proved that altitudes are equal to each other. Further we can proove that the perpendiculars is equal to the altitudes on either of the equal angles as (angle opposite to equal sides are equal)

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