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

Similarly,

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.

Now,

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]

But

∠AMF+∠AME** **=

⇒2∠AME=

Hence, ∠AMF =∠AME =

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