Question

If F is not considered as mid point of AC of ∆ABC from above theorem, drawing perpendicular on AC from O prove that the perpendicular passes through mid point of AC.


Hey guys pls help me out this​

Answer 1

Answer:

R.E.F image

Given : △ ABC is isosceles with AB=AC ,E and F are the mid-points of BC, CA and AB

To prove: AD⊥EFand is bisected by t

construction: Join D, F and F

Proof: DE∣∣AC and DE=

2

1

AB

and DF∣∣Ac andDE=

2

1

AC

The line segment joining midpoints of two sides of a triangle is parallel to the third side and is half of it

DE = DF (∵AB=AC) Also AF=AE

∴AF=

2

1

AB,AE=

2

1

AC

∴DE=AE=AF=DF

and also DF∣∣ AE and DE∣∣AF

⇒ DEAF is a rhombus.

since diagrams of a rhombus bisect each other of right angles

∴AD⊥EF and is bisected by it