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Consider the following diagram:

enter image description here

$AB+AD=DE$, $\angle BAD= 60$, and $AE$ is $6$. How do we find the area of the triangle $ABC$?

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1  
Is $AE$ the diameter of the circle? –  Sumit Bhowmick Nov 7 '12 at 5:31
1  
@Sumit Yes, it is. –  George Krasilnikov Nov 7 '12 at 5:38
    
please find the answer. –  Sumit Bhowmick Nov 7 '12 at 14:08
    
@Sumit is what you found not the answer? –  George Krasilnikov Nov 7 '12 at 14:13

1 Answer 1

up vote 1 down vote accepted

enter image description here

From the above picture,

$x+y=z$ , $ y+z=6$ and $\frac x {y+z} = \cos 60^\circ = \frac 1 2$

After calculation, $x=3$, $y=\frac 3 2$ and $z = \frac 9 2$

$\angle ABE = 90^\circ$ and $\angle BAD = 60^\circ$

So, $\angle AEB = 30^\circ = \angle ACB$ (properties of a circle)

Now, $\cos 60^\circ = \frac {x^2 + y^2 - w^2}{2xy} = \frac 1 2$

After calculation, $w=DB=\frac {3 \sqrt 3}{2}$

$$\cos \angle ADB=\frac{y^2+w^2-x^2}{2yw}=\frac{(\frac 3 2)^2+(\frac {3 \sqrt 3}{2})^2-3^2}{2 \frac 3 2\frac {3 \sqrt 3}{2}}=0$$

So, $\angle ADB = 90^\circ$ and $\angle ABD = 30^\circ = \angle ACB$

So, $\triangle ABC $ is an isosceles triangle.

And, $CD = BD = \frac {3 \sqrt 3}{2}$

So, $\triangle ABC = \frac 1 2 \cdot \frac 3 2 \cdot (\frac {3 \sqrt 3}{2} + \frac {3 \sqrt 3}{2}) = \frac {9 \sqrt 3}{4}$

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@George Krasilnikov please find the answer. –  Sumit Bhowmick Nov 7 '12 at 8:45
    
@Sumit... is what you gave not the answer? –  George Krasilnikov Nov 7 '12 at 14:11
    
@GeorgeKrasilnikov, yes this is my answer. –  Sumit Bhowmick Nov 8 '12 at 4:48

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