Let △ABC be an isosceles triangle with AB=AC and let D, E and F be the midpoints of
BC, CA and AB respectively. Show that AD⟂ FE and AD is bisected by FE.
Answers
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
hope this is helpful to you
Answer:
hey guys
it's me Sahana and here use your answer for the above question
Step-by-step explanation:
ANSWER
R.E.F image
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