Question

In an equilateral ∆ABC,AD is altitude from A on BC,.Prove that 3AB sq=4AD sq

Answer 1

Since ABC is an equilateral triangle, AB = BC = AC

AD is the altitude on BC so AD ⊥ BC.
Altitude = Median = Perpendicular bisector

BD = DC = BC/2
Since, AB = BC = CA

So in place of BC,we can plug AB
∴ BD = DC = AB/2 -----(i)

In ΔABD,
AB² = AD² + BD²
AB² = AD² + (AB/2)²   ----From (i)
AB² = AD² + AB²/4
AD² = AB² - AB²/4
AD² = (4AB² - AB²)/4
AD² = 3AB²/4
4AD² = 3AB²

So , 3AB² = 4AD²

Hence Proved!

Hope This Helps You!

Answer 2

Hey there !


ABC is an equilateral triangle.
So ,
AB = BC = AC

AD is the altitude from A on BC and
AD ⊥ BC.
Here ,

Altitude = Median = Perpendicular bisector

BD = DC = BC/2

AB = BC
So , replacing BC with AB

BD = DC = AB/2 -----> [ 1 ]

In ΔABD,

Pythagoras theorem:-

AB² = AD² + BD²

From [ 1 ]
AB² = AD² + (AB/2)²

AB² = AD² + AB²/4

Taking AB to the left side,
AD² = AB² - AB²/4

lcm = 4
AAD² =[ 4AB² / 4] - [AB²/4]

AD² = 3AB²/4

4AD² = 3AB²

Therefore ,
3AB² = 4AD²
------> Proved !!