In triangle abc d e and f are the midpoints of the sides ab ac and bc. Prove that de and ef will bisect each other
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Answer:
dearfriend
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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
DE = 1/2 AB. ( by mid point of theorem )
Similarly,
DF II AC and
DF =1/2AC
therefore AB = AC
=> 1/2AB= 1/4 AC..........1
so DE = DF
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 = 180
⇒2∠AME=180
Hence, ∠AMF =∠AME =