30.ABC be an isosceles triangle with AB = AC and let D,F,E are the mid-points of BC,CA and AB
respectively. Show that AD perpendicular to EF and AD bisector of EF.
Answers
Given : △ ABC is isosceles with AB=AC ,E and F are the mid-points of BC, CA and AB
To prove: AD⊥EF and is bisected by t
construction: Join D, F and F
Proof: DE∣∣AC and DE=
2
1
AB
and DF∣∣Ac and DE=
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
solutio
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Question :
ABC be an isosceles triangle with AB = AC and let D,F,E are the mid-points of BC,CA and AB respectively. Show that AD perpendicular to EF and AD bisector of EF.
Given :
ΔABC is isosceles triangle with AB = AC.
D, E and F are the mid points of BC, CA and AB respectively.
To prove :
AD ⊥ EF and bisected by it.
Construction : Join D, E and F
Proof :
DE || AB and DE = ¹/₂ AB
and
DF || AC and DF = ¹/₂ AC
[ ∵ Line segment joining midpoints of two sides of a triangle is parallel to the third side and half of it ]
DE = DF [ ∴ Ab = AC ]
And
AF = AE [ ∴ AF = ¹/₂ AB, AE = ¹/₂ AC ]
∴ DE = AE = AF = DF and also DF || AE and DE || AF
⇒ DEAF is a rhombus
Since, diagonals of a rhombus bisect each other at right angles.
∴ AD ⊥ EF and is bisected by it.
