Question

ABC is an isoceles triangle with AB=AC.D,E and F are mid points of the sides BC,AB and AC respectively. Prove that the line segment AD is perpendicular to EF and is bisected by it.
plzzz solve this....fast..

Answer 1

$\textbf{\huge{\underline{ Hello Mate! }}}$

Given : AB = AC and D, E and F are mid points in BC, AB and AC respectively.

To Prove : AD is perpendicular to EF and is bisected by it

To construct : Join DE and DF.

Proof : Since EF are formed by joing mid points of triangle then EF || BC.

And since parallel lines are at 90° angle hence AD is perpendicular to EF.

Now, DF || AB and DF = 1 AB / 2

Since 1 AB / 2 = BE

DF = BE

Now, AB = AC

1 AB / 2 = 1 AC / 2

BE = AF

DF = AF

Now, taking triangle AOF and DOF.

DF = AF ( Proved above )

OF = OF ( Common )

angle AOF = ange DOF ( 90° each )

Hence triangle AOF ~ DOF by RHS congruency. Thus, AO = OD.

Hence EF bisect AD.

$\textbf{\large{ Q.E.D }}$

$\textbf{ Have great future ahead!}$

Answer 2

$dear \: friend \:$


here is your answer OK ☺☺☺☺


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 =90