let ABC be an isosceles triangle in which AB = AC if D, E, F be the mid-points of - Maths - Quadrilaterals

Question

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

Gursheen kaur. · Accepted Answer

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

$D E = \frac{1}{2} A B [B y M i d - p o int T h e o r e m]$

Similarly,

DF II AC and

$D F = \frac{1}{2} A C ∴ A B = A C \Rightarrow \frac{1}{2} A B = \frac{1}{2} A C ............. (I) \Rightarrow D E = D F ......... (i)$

Now,DE II FA and DE = FA [ $∵$ DE II AB and DE = $\frac{1}{2} A B = F A$ ]

⇒ 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 = $180^{0}$

⇒2∠AME= $180^{0}$

Hence, ∠AMF =∠AME = $90^{0}$

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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.