In an equilateral triangle of side $3\sqrt{3}$ 𝑐𝑚, find the length of the altitude.

                                                     (OR)

ABC is an isosceles triangle in which AB=AC which is circumscribed about a circle as shown in the figure. Show that BC is bisected at the point of contact.

It is given that the sides of an equilateral triangle 𝐴𝐵𝐶 is $3\sqrt{3}$ 𝑐𝑚. 

I.e 𝐴𝐵 = 𝐵𝐶 = 𝐶𝐴 =  $3\sqrt{3}$𝑐𝑚

Let 𝐴𝐸 = ℎ (altitude) Since altitude bisects the base 𝐵𝐸 = $\frac{1}{2}$𝐵𝐶

$\begin{array}{r}BE=\frac{3\sqrt{3}}{2}\mathrm{cm}\\ A{B}^2=B{D}^2+A{D}^2\\ (3\sqrt{3}{)}^2=h^2+{\left(\frac{3\sqrt{3}}{2}\right)}^2\\ 27=h^2+\frac{27}{4}\\ h^2=27-\frac{27}{4}\\ h^2=\frac{27\times 4-27}{4}\\ h^2=\frac{81}{4}\\ h=\sqrt{\frac{81}{4}}\\ h=\frac{9}{2}\\ h=4.5\mathrm{cm}\end{array}$

Hence, the length of altitude is 4. 5 𝑐𝑚

                                                   (OR)

It is given that in an isosceles triangle AB=AC which is circumscribed about a circle.

As tangents drawn from an external point to a circle are equal in length. 

So, we get 

𝐴𝑃 = 𝐴𝑄 (tangents from A) 

𝐵𝑃 = 𝐵𝑅 (tangents from B) 

𝐶𝑄 = 𝐶𝑅 (tangents from C) 

It is given that ABC in an isosceles triangle with sides 

𝐴𝐵 = 𝐴𝐶 

𝐴𝐵 − 𝐴𝑃 = 𝐴𝐶 − 𝐴𝑃 

𝐴𝐵 − 𝐴𝑃 = 𝐴𝐶 − 𝐴𝑄 

𝐵𝑅 = 𝐶𝑄 

Since 𝐵𝑅 = 𝐶𝑄 that implies BC is bisected at the point of contact.