let ABC be an isosceles triangle in which AB = AC. if D, E, F be the mid-points of the sides BC, CA and AB respectively, show that the segment AD and EF bisect each other at right angles.
Given: ABC be an isosceles triangle in which AB = AC in which join the mid-points D ,E,F on BC,AC&AB.
Proof: Let AD intersect FE at M.
Join DE and DF.
Now, D and E being the mid points of the sides BC and CA resp.
DE II AB and
DF II AC and
Now,DE II FA and DE = FA [ DE II AB and DE = ]
⇒ DEAF is a IIgm
∴AM=MD & FM =ME.........(II)
⇒ DEAF is a rhombus.[ DE =DF, from (i), DE =FA and DF = EA]
But, the diagonals of a rhombus bisect each other at right angles.
∴ AD ⊥FE and AM = MD.
Hence, AD ⊥FE and AD is bisected by FE.
In Δ AFM and Δ AEM,
⇒AF = AE [from (I)]
⇒AM=AM (common side)
⇒FM =ME [from (II)]
By SSS congruence
Δ AFM Δ AEM
∠AOF =∠AOE [By cpct]
Hence, ∠AMF =∠AME =
Draw triangle ABC. In which join the mid-points D ,E,F on BC,AC&AB. Now draw AD on BC. mark "O" where EF & AD intersect .
- Since it's given that AB =AC, & E & F are the midpoint of AB & AC , thus 1/2AB =1/2AC (1). thereefore AF=AE.
- Now DEAF is a //gm (since AF II ED & AF =ED)
- And we know that diagonal of //gm bisect each other , therefore AO =OD & FO =OE(2)
- Now consider triangle AFO & AEO, in this AF = AE (fron 1)& AO=AO (common side) FO =OE (from 2). Thus By SSS the triangle are congruent By CPCT angle AOF =AOE.
- But since angle AOF+AOE = 180 , =2AOE=180
therefore angle AOF =AOE = 90