In equilateral △ ABC, AD ⊥ BC and BC = x cm. Find, in terms of x, the length of AD.

In △ ABD and △ ACD,

⇒ ∠ADB = ∠ADC (Both equal to 90°)

⇒ AD = AD (Common side)

⇒ AB = AC (Since, ABC is an equilateral triangle)

∴ △ ABD ≅ △ ACD (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

∴ BD = CD = $\dfrac{BC}{2} = \dfrac{x}{2}$ cm.

In right-angled triangle ABD,

By pythagoras theorem,

⇒ (Hypotenuse)² = (Perpendicular)² + (Base)²

⇒ AB² = AD² + BD²

⇒ x² = AD² + $\Big(\dfrac{x}{2}\Big)^2$

⇒ AD² = x² - $\Big(\dfrac{x}{2}\Big)^2$

⇒ AD² = $x^2 - \dfrac{x^2}{4}$

⇒ AD² = $\dfrac{4x^2 - x^2}{4}$

⇒ AD² = $\dfrac{3x^2}{4}$

⇒ AD = $\sqrt{\dfrac{3x^2}{4}} = \dfrac{\sqrt{3}x}{2}$ cm.

Hence, AD = $\dfrac{\sqrt{3}x}{2}$ cm.