Question

Prove that?

Answer 1

Let `AD`, the perpendicular drawn from A to the angle bisector `BD` of `/_ABC` be produced to intersect `BC` at E.

Now in `DeltaABD and DeltaEBD` we have

• `/_ABD=/_EBD " given"`

• `/_ ADB=/_ EDB,` since BD perpendicular to AE

• Side `BD` is common

So the triangles are congruent following Rigt angle Side Angle rule.
Hence `AD=DE`. This means D is the mid point of `AE`. So if a line is drawn through D parallel to EC or BC it will biseect AC.