Question

ABC is an isosceles triangle with AB = AC and let D, F, E be the mid-points of BC, CA and AB respectively. Show that AD is perpendicular to EF and AD bisects EF.

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Answer 1

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Answer 2

$\large\underline\bold{Given:}$

ABC is an isosceles triangle, where AB = AC

D is the mid point of BC

F is the mid point of CA

E is the mid point of AB

$\large\underline\bold{To~Prove:}$

AD is perpendicular to EF i.e. angle AOE = angle AOF = 90°
AD bisects EF i.e. EO = OF

$\large\underline\bold{Proof:}$

We know that the median from vertex of an isosceles triangle divides it into two congruent triangles so,
angle BAD = angle CAD ------------------ ( 1 )

AB = AC [ Given ]
Multiplying both side by 1 / 2,
AB / 2 = AC / 2
AE = AF [ Given, E is the mid point of AB and F is the mid point of AC ] --------------- ( 2 )

Now,
In ∆AOE and ∆AOF,
AO = AO [ Common ]
angle BAD = angle CAD [ Proved in equation ( 1 ) ]
AE = AF [ Proved in equation ( 2 ) ]
Therefore,
∆AOE congruent to ∆AOF by SAS postulate

Now by CPCTC,
angle AOE = angle AOF and EO = OF

So,
AD bisects EF

Also,
angle AOE + angle AOF = 180° [ Linear Pairs ]
2( angle AOE ) = 180°
angle AOE = angle AOF = 180° / 2 = 90°

So,
AD is perpendicular to EF