Question

ABC is isoceles triangle with AB=AC. D,E,F, are midpoints of BC,AB,AC respectively. Then show that AD is perpendicular to EF.

Answer 1

I think this is the answer

Answer 2

given- ∆ABC with mid points D,E,F of sides BC,AB,AC respectively.

to prove- AD is perpendicular to EF.

proof- in ∆ ABC, AD forms the median of the ∆, so BC=DC.
in ∆ABD and ∆ ACD:
AB=AC (given)
BD=DC (AD is median)
AD=DA (common)
so ∆ABD is congruent to ∆ACD (by SSS congruence criterion)

so angle ADB= angle ADC (by cpct)

angle ADB+ ADC= 180° (by linear pair)
angle ADB+ADB= 180° (since ADB=ADC)
2angle ADB = 180°
angle ADB = 180÷2
angle ADB = 90°

so AD is perpendicular to BC.

now since E is mid point of AB and F is mid point of AC, so EF is parallel to BC and is equal to half of BC by mid point theorem.

so angle APF is also 90° by corresponding angles (EF||BC and AD is transversal)

so AD is also perpendicular to EF.
hence proved.