# How to solve this geometry problem involving circles and tangents?

In triangle ABC the angle at A is 60 and the inscribed circle touches AB at the point D. If AD = 5 and DB = 3, find the length of BC.

I know BD = BE since they are tangents and AD = AF but this is the furthest I've reached into solving this problem but I can't think of anything else to do.

Let $O$ be the center of the inscribed circle. Then $\Delta ADO$ forms a 30-90-60 $\Delta$. Hence you get the radius $r$ of the inscribed circle. Use this in $\Delta OBE$ to get angle $B$ and hence angle $C$ and hence $EC$.

• how do I get the radius $r$ of the circle? if you are saying that $ΔADO$ forms a 30-90-60 then how do I proceed from there? – Aspiring Mathlete Mar 9 '14 at 16:37
• $AD$ is given and you know $\angle OAD$. Using trigonometric formula for $\tan$ you can calculate $r$. – Sandeep Thilakan Mar 10 '14 at 4:58
• But I do not know the value for AO nor do I know the value for OD. Also, I am not allowed to use a calculator. – Aspiring Mathlete Mar 10 '14 at 15:43
• $\tan \angle OAD = \tan 30 = \frac{OD}{AD}$ Therefore $r = \frac{5}{\sqrt{3}}$ – Sandeep Thilakan Mar 11 '14 at 5:43
• OK. I know now that $OE$ = $r$. Now how do I proceed? I cannot get the value of angle B without a calculator and I am not allowed to use one. Also, the memorandum says that that BC = 13, but I do not know how to get that answer. Please can you show me how you get the value of BC? – Aspiring Mathlete Mar 11 '14 at 15:47

there is property of circle construction :Inscribe a Circle in a Triangle

Steps:

Bisect one of the angles
Bisect another angle
Where they cross is the center of the inscribed circle


now if you bisect one angle ,namely angle $A$,but this also would be right triangle,because The angle between the tangent and radius is 90°,example

and you can easily proceed from here

The tangents CF and CE are also equal - call their value $x$, then apply the cosine rule. The extent of the trigonometry required is knowing that $\cos(60) = \frac{1}{2}$.