Finding the Relationship Between Altitude and Side in an Equilateral Triangle

In an equilateral triangle, all three sides are equal in length, and all three angles are equal to 60 degrees. Let the equilateral triangle be ABC, where AB = BC = CA.

We are given that AD is the altitude from vertex A to the side BC, and AD is perpendicular to BC (\(AD \perp BC\)).

A key property of an equilateral triangle is that the altitude from a vertex to the opposite side also bisects that side. Therefore, D is the midpoint of BC.

So, we have \(BD = DC = \frac{1}{2} BC\).

Since ABC is an equilateral triangle, \(BC = AB\). Thus, \(BD = \frac{1}{2} AB\).

Now, consider the triangle ADB. Since \(AD \perp BC\), angle ADB is a right angle (\(\angle ADB = 90^\circ\)). Triangle ADB is a right-angled triangle.

We can apply the Pythagorean theorem to the right-angled triangle ADB. The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In triangle ADB:

• Hypotenuse = AB
• One side = AD (altitude)
• Other side = BD (half of the base)

According to the Pythagorean theorem:

\(AB^2 = AD^2 + BD^2\)

We know that \(BD = \frac{1}{2} AB\). Substitute this into the equation:

\(AB^2 = AD^2 + \left(\frac{1}{2} AB\right)^2\)

\(AB^2 = AD^2 + \frac{1}{4} AB^2\)

Now, we need to rearrange the equation to find the relationship between \(AD^2\) and \(AB^2\). Subtract \(\frac{1}{4} AB^2\) from both sides:

\(AB^2 - \frac{1}{4} AB^2 = AD^2\)

\(\frac{4AB^2 - AB^2}{4} = AD^2\)

\(\frac{3AB^2}{4} = AD^2\)

Finally, multiply both sides by 4 to get rid of the fraction:

\(3AB^2 = 4AD^2\)

Or, writing \(AD^2\) on the left side:

\(4AD^2 = 3AB^2\)

This equation shows the relationship between the square of the altitude (\(AD^2\)) and the square of the side (\(AB^2\)) in an equilateral triangle.

Comparing this result with the given options, we find that the statement \(4AD^2 = 3AB^2\) is true.