Question

A triangle ABC has Angle B = C. prove that the perpendiculars from the mid point of BC TO AB are equal

Answer 1

Let there be a traingle where AP is perpendicular to BC and CQ is perpendicualar to AB...
NOW.. In traingle ABC and BCQ
And since AP and &CQ are perpendicualrs so.....
SO TRIANGLE APC~CQB... BY AA similarity rule
SO... .
AP=CQ (cpst) i.e. congruent parts of simailar triangles

HOPE IT HELPS