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


∠AOF =∠AOE [By cpct]




Hence, ∠AMF =∠AME =




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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 .

  1. Since it's given that AB =AC, & E & F are the midpoint of AB & AC , thus 1/2AB =1/2AC (1). thereefore AF=AE.
  2. Now DEAF is a //gm (since AF II ED & AF =ED)
  3. And we know that diagonal of  //gm bisect each other , therefore AO =OD & FO =OE(2)
  4. 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.
  5. But since angle AOF+AOE = 180 , =2AOE=180

therefore angle AOF =AOE = 90

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