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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.
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Solution
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.
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