Question

ΔABC is an isosceles triangle with AB = AC. D, E and F are the mid-points of BC, AB and AC respectively. Prove that the line segment AD is perpendicular to EF and is bisected by it.

Answer 1

Since the segment joining the mid-points of two sides of a triangle is parallel to third side and is half of it,
Therefore,
DE || AB, DE = `(1)/(2)"AB"`

Also,

DF || AC, DF = `(1)/(2)"AC"`

But AB = AC

⇒ `(1)/(2)"AB" = (1)/(2)"AC"`

⇒ DF = DE    ........(i)

DE = `(1)/(2)"AB"`

⇒ DE = AF   ........(ii)

And DF = `(1)/(2)"AC"`

⇒ DF = AE  ........(iii)

From (i), (ii) and (iii)
DE = AE = EF = DF
⇒ DEAF is a rhombus.
⇒ Diagonals AD and EF bisect each other at right angles.
⇒ AD perpendicular to EF and AD is bisected by EF.