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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!

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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?

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

$\angle ABE=30^o$

$\angle ABD=90^o$

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:

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

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