# Geometry question involving the length of a chord

Two chords $$AB$$ and $$AC$$ are drawn inside a circle with diameter $$AD$$. The angle $$BAC = 60$$, $$AB = 24cm$$, $$EC = 3cm$$, and $$BE$$ and $$AC$$ are perpendicular. What is the length of the chord $$BD$$?

Here's what I've tried:

$$ABE = 30$$ which implies $$AE = 12cm$$ and therefore $$BE = 12\sqrt3cm$$ and $$BC = 21cm$$. Call the intersection of $$EB$$ and $$AD$$ point $$O$$. So we have $$AEO$$ is similar to $$ACD$$ but I don't know where to go from here.

Thank you so much in advance!

You are nearly there!

Note that angles CBD and CAD are equal. Angles BAD and EBC are then easily proved to be equal.

The right-angled triangle ABD is now similar to the triangle BEC of which you know all dimensions. Over to you?

• I'm so sorry but could you please say how you got $CBD = CAD$ (the angles) – Borna Ahmadzade Oct 9 '19 at 16:29
• This is the 'angle in a sector' result. CBD and CAD are both angles on chord CD and are therefore equal. – S. Dolan Oct 9 '19 at 20:05

Another approach:

$$\angle ABE=30^o$$

$$\angle ABD=90^o$$

because it is opposite to diameter AD.Therefore:

$$\angle OBD=90-30=60^o$$

$$DF||EC$$. SO :$$DF=EC=3$$

In right angled triangle BFD we have:

$$BD=\frac{3}{Sin(60)}=\frac{3}{\sqrt 3/2}=2\sqrt 3$$