Answer:
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=
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=
AB,AE=
∴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
solution
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Answers & Comments
Answer:
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
solution