A geometry problem by Dimple Maniar

To find the area of ΔCDE we must know <CDB, which is equal to <BAC. We can calculate the <BAC=x, as follow, (180 – x)/2 – 15 + x = 90, x= 30. Look at ΔCDE, CD/sin135 =CE/sin 30, CE=2√3/sin135. The area of ΔCDE = ½ CE . CD sin 15 = 12 sin 15/sin135 = 4.392

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Let $\alpha=\angle CBD$ , such that $\angle BAC = \angle BDC = 90^{\circ}-\alpha.$ In the isosceles triangle $\triangle ABC$ ,

$90^{\circ} -\alpha + 2(\alpha + 15^{\circ} ) = 180^{\circ}$ , therefore, $\alpha = 60^{\circ}$ . Thus, $\triangle BOC$ is an equilateral triangle, and $BC=4.$

We also have $\angle BDC = 90^{\circ}-\alpha = 30^{\circ}$ .

In triangle $\triangle COD$ , $\overline{CE}$ is the angle bisector of $\angle OCD$ . According the angle bisector theorem, $\dfrac{DE}{OE}= \dfrac{CD}{OC}.$ Since $CD = \sqrt{BD^2-BC^2} =4\sqrt{3}$ , we have $DE = 6-2\sqrt{3}$ .

Since $\angle BDC =30^{\circ}$ , the altitude from $E$ to $CD$ is $DE/2=3-\sqrt{3}.$ Therefore, the triangle area $[CDE] = \dfrac{(3-\sqrt 3)\cdot 4\sqrt{3}}{2} = \boxed{4.392}.$

I've got a bit long calculation part but its not very complicated.

$Angle BAC= Angle BDC=y$ (Since chord subtends equal angles in a segment)

$Angle ABC= Angle ACB= x$ (Since $AB=AC$ )

In $\bigtriangleup ABC and \bigtriangleup BDC$ , by equating the two values of $y -$

$180-2x=180-(x-15)-90$

$x=75$

Now,

$\sin DBC= \frac{DC}{BD}$

$\sin(x-15)= \frac{DC}{8}$

$\sin60= \frac{DC}{8}$

$DC= 4\sqrt{3}=\boxed{6.9282}$

By sine rule,

$\frac{\sin135}{DC}=\frac{\sin15}{DE}=\frac{\sin30}{CE}$

Therefore,

$DE=\frac{\sin15 \times DC}{\sin135}=\boxed{2.5358}$

$CE=\frac{\sin30 \times DC}{\sin135}=\boxed{4.8989}$

Now put the values in Heron's Formula

$Area of \bigtriangleup CDE= \sqrt{7.18145(4.64565)(2.28255)(0.25325)}$

$Area of \bigtriangleup CDE= 4.3915 \approx \boxed{4.392}$

Angle BCD= 90 Degrees ( Angle in Semi - Circle) Also Angle ABD = Angle ACD = 15 Degrees ( Angle subtended by the same arc AD) Therefore Angle ACB = 90 -15 = 75 Degrees => Angle DBC = 60 Degrees => Angle BDC = Angle BAC = 30 Degrees ( Angle subtended by the same arc BC) => Angle CED = 135 Degrees

Now consider the Right triangle BCD. CD = 8 * (sin 60) = 6.928 --------------- (1)

Apply Sine Rule to the triangle CED to determine EC

EC = (CD) * (sin 30) / (sin 135) = 4.89 --------------- (2)

Therefore Area of CED = (0.5) * (CD) (EC) (sin ECD) = (0.5) (6.928) (4.89)*(sin 15) = 4.39 Sq Units