Question

Prove that the altitude from the vertex of an isosceles triangle to it's opposite base bisects the base. ​

Answer 1

Answer:

let us take a triangle ABC and draw an altitude CD

being an isosceles triangle

in ΔACD and ΔCDB

CD=CD (common )

CA=CB ( equal opposite sides)

∠ A=∠ B ( angle opposite to opposite sides are also equal )

therefore : ΔACD and ΔCDB are congruent by ASA rule

so , AD=DB by CPCT

So, its true