Question

BE and CF are two equal altitudes of a ∆. Prove that the ∆ is

isosceles.​

Answer 1

Step-by-step explanation:

BE is altitude,

So, <AEB = <CEB =90°. --------(1)

Also, CF is an altitude,

So, <AFC = <BEC =90°. ---------(2)

BE = CF. -----------(3)

TO PROVE: ∆ABC is isosceles

In ∆BCF and. ∆CBE

<BFC = <CBE. (90°)

BC. =. CB. (common)

FC = EB (from(3))

∆BCF =~ ∆CBE. (By RHS)

Therefore, <FBC = <ECB (CPCT)

So, <ABC = <ACB

AB = AC. (sides opposite to equal angle is equal)

So, ∆ABC is an isosceles triangle..

please mark it as brilliant.