ABC is an isosceles triangle with AB = AC = 2a and BC = a. If AD ⊥ BC, find the length of AD.

Given: AB = AC = 2a, BC = a and AD ⊥ BC.

Since Δ ABC is isosceles with AB = AC, the perpendicular from A to BC will bisects BC, meaning:

BD = DC = $\dfrac{\text{BC}}{2} = \dfrac{a}{2}$

In right-angled triangle ADB,

AB = 2a

BD = $\dfrac{a}{2}$

Using the Pythagorean theorem,

AB² = AD² + BD²

⇒ (2a)² = AD² + $\Big(\dfrac{a}{2}\Big)$ ²

⇒ 4a² = AD² + $\dfrac{a^2}{4}$

⇒ AD² = 4a² - $\dfrac{a^2}{4}$

⇒ AD² = $\dfrac{16a^2}{4} - \dfrac{a^2}{4}$

⇒ AD² = $\dfrac{15a^2}{4}$

⇒ AD = $\sqrt{\dfrac{15a^2}{4}} = \dfrac{a\sqrt{15}}{2}$

Hence, the length of the perpendicular from A to BC is $\dfrac{a\sqrt{15}}{2}$ units.