9.5 Cyclic Quadrilaterals

Step 1

$\angle BDC = \angle BAC = 30^\circ$ (Angles in same segment).

Step 2

In $\Delta BCD$ , $\angle B + \angle D + \angle C = 180^\circ$ . $\angle DBC = 70, \angle BDC = 30$ .

Step 3

$70 + 30 + \angle BCD = 180 \implies \angle BCD = 80^\circ$ .

Step 4

If $AB=BC$ , $\Delta ABC$ is isosceles. $\angle BAC = \angle BCA = 30^\circ$ .

Step 5

$\angle BCD = 80^\circ$ . We know $\angle BCA = 30^\circ$ .

Step 6

$\angle ECD = \angle BCD - \angle BCA = 80 - 30 = 50^\circ$ .

Final Answer

80°, 50°

Operation

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Radius (r)

Chord Length (c)

Dist from Center (d)

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