Question

ABC is an equilateral triangle. AD is perpendicular on the side BC, let us Prove that, AB2 +BC2 +CA2 = 4AD2.​

Answer 1

Answer:

Given :

• ABC is an equilateral triangle. AD is perpendicular on the side BC .

Prove That :

• AB²+ BC²+ CA² = 4AD².

Proof :

In the equilateral ∆ABC, AD $\perp$ BC

Then, BC = DC

So, ∆ABD and ∆ACD both are right angle.

∴ AB² = AD²+ BD² and OA² = AD²+ DC²

∴ AB² = AD²+ BD² = AD²+ DC² [ $\because$ BD = DC ]

Now, AB²+ BC²+CA²

=AD²+ DC²+ (2DC)²+ AD²+ DC² [ $\because$ BC = DC ]

= AD² +DC²+ 4DC²+ AD² + DC²

= 2AD² + 6DC²

Again, AD is the hight of ∆ABC.

∴ AD = $\dfrac{\sqrt{3}}{2}$.AC = $\dfrac{\sqrt{3}}{2}$.(2DC) [ $\because$ AB = BC = CA = 2DC ]

⇒ $\sqrt{3}$ .DC

∴ AD² = 3DC²

⇒ DC² = $\dfrac{1}{3} A{D}^{2}$

∴ AB² + BC² + CA²

↦ 2AD² + $6.\dfrac{1}{3}$AD²

↦ 2AD² + 2AD²

➦ 4AD²

Hence, AB²+ BC²+ CA² = 4AD². (PROVED)

Answer 2

Answer:

$</p><p>\color{maroon}\mathbb{HOPE IT WILL HELP YOU MY DEAR FRIEND AND PLEASE GIVE ME MULTIPLE THANKS}$

Given AD⊥ BC

D=90˚

Proof:

Since ABC is an equilateral triangle,

AB=AC=BC

ABD is a right triangle.

According to Pythagoras theorem,

AB

2

=AD

2

+BD

2

BD=1/2BC

AB

2

=AD

2

+(1/2BC)

2

AB

2

=AD

2

+(1/2AB)

2

[∵BC=AB]

AB

2

=AD

2

+1/4AB

2

AB

2

=(4AD

2

+AB

2

)/4

4AB

2

=4AD

2

+AB

2

4AD

2

=4AB

2

–AB

2

4AD

2

=3AB

2

Hence proved.