As easy as ABC

Let $\angle ABC = x$ and $\angle AIB = \angle EID = y$ . Note that $\angle BAI = \frac{\angle BAC}{2} = 23.5^\circ$ .

In $\triangle AIB$ ,

$\angle ABI + \angle BAI + \angle AIB = 180^\circ$

$\Rightarrow \frac{x}{2} + 23.5^\circ + y = 180^\circ$

$\Rightarrow \frac{x}{2} + y = 156.5^\circ$ _ _ _ _ _(1)

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Quadrilateral IDCE is inscribed in a circle.

So, $\angle EID + \angle ECD = 180^\circ$

$\angle ECD = 180^\circ - y$

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In $\triangle ABC$ ,

$\angle BAC + \angle ABC + \angle ACB = 180^\circ$

$\Rightarrow 47^\circ + x + (180^\circ - y) = 180^\circ$

$\Rightarrow y - x = 47^\circ$ _ _ _ _ _(2)

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Subtract (2) from (1).

$\frac {3x}{2} = 109.5^\circ$

$\Rightarrow x = \angle ABC = \boxed {73^\circ}$

B/2 + A/2 = C (Concyclicity & external angle) or 90° - C/2 = C or C =60°; hence, B=73°.

Denote $\angle BAC = \alpha$ , $\angle ABC = \beta$ , $\angle BCA = \gamma$ , then the fact that quadrilateral $CEID$ is cyclic means that $\angle AIB = \angle EID = 180^{\circ} - \gamma = \alpha + \beta$ Also $\angle IAB = \frac{\alpha}{2}$ and $\angle IBA = \frac{\beta}{2}$ , hence in triangle $\triangle ABI$ we get $\alpha + \beta + \frac{\alpha}{2} + \frac{\beta}{2} = 180^{\circ}$ i.e. $\beta = 120^{\circ} - \alpha = 120^{\circ} - 47^{\circ} = \boxed{73^{\circ}}$

As I lies on circumcircle of ECD, IECD is a cyclic quadrilateral so EIA = 180-C. C = 133-B (By angle sum property of a triangle) EIA = 47+B. AIB = 180-(47+B)÷2 = EIA = 47+B. SOLVE THIS EQUATION AND YOU WILL GET ABC = 73.

angle AOB = 90+C/2. Given 90+C/2 + C = 180..Thus C = 60. Hence B=73

Observe that $IDCE$ is a cyclic quadrilateral, $\Rightarrow$ $\angle DIE = 180^ \circ - \angle C$

$\Rightarrow \angle BIA = 180^ \circ - \angle C$

$\Rightarrow \angle A/2 + \angle B/2 + 180^ \circ - \angle C = 180^ \circ$

$\Rightarrow (90^ \circ -\angle C/2 ) = \angle C \Rightarrow \angle C = 60^ \circ .$

$\Rightarrow \angle B = 180^ \circ - 60^ \circ - 47^ \circ = 73^ \circ .$

With the condition of the problem, CDIE is ciclic, then Let be <BCA=2a => <EIA=2a ( ) but by the other hand, <B=133-2a => <EBA=133/2-a and by the triangule IBA, <EIA=133/2+47/2-a ( ), by * and * , 90-a=2a => a=30 => <B=133-2a=133-60=73.

angle EID + angleECD =180(1) (EICD IS A CYLIC QUAD.) IN TRIANGLE ABI 1/2 (A+B)+EID=180 = A +B +2 EID =360 (2) Equate (1) and (2) You will get the value of EID then find the value of C AND B