Question

E) If ∆DEF is equilateral triangle & seg DP is perpendicular
side EF. E-P-F then prove that ,DP²= 3EP².
​

Answer 1

If AN⊥BC

Then △ANB will be a right angle △ ,

Using Pythagorean theorem, we get

(AN)

2

+(BN)

2

=(AB)

2

On subtracting (BN)

2

from both side

(AN)

2

=(AB)

2

−(BN)

2

Since triangle ABC is equilateral, perpendicular segment AN to BC bisects line segment BC,

Therefore, AB=2BN then

(AN)

2

=(2BN)

2

−(BN)

2

(AN)

2

=4(BN)

2

−(BN)

2

(AN)

2

=3(BN)

2

Hence proved.