Question

in the figure, AD is the median of ∆ ABC. BE And CF are perpendiculars from B and C and on a AD and AD produced. prove that BE= CF.​

Answer 1

ΔBCE & ΔBFC,

1. BC=CB ( Common side)

2. angle BEC= angle CFB=90°( perpendiculars)

3. BF= CE ( mid pt. theorem)

therefore, by S.A.S, ΔBCE congruent to ΔBFC

By C.P.C.T , BE=CF (proved)