# Cyclic Quadrilateral QuestionABCD is a cyclic quadrilateral in which AD=DC=CB. Prove that: a) angle ADC = angle DCB b) DC is parallel to AB

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Draw this cyclic quadilateral and mark ABCD in order.

Draw the two diagonals (BC and AD)

Now in ADC Triangle.

DAC angle = DCA angle (since AD = DC)

Let's call this two angles as x,

therefore, DCA angle = DAC angle = x

Therefore ADC angle = 180 - 2x (Angles in a triangle)

But in a cyclic quadilateral the opposing angles are complementary,

ADC angle + ABC angle = 180

180 -2x + ABC angle = 180

ABC angle = 2x

Conside the circle segment AD,

Now DCA angle = DBA angle (Angles in same segment)

Since DCA angle = x

Therefore DBA angle = x

But

ABC angle = DBA angle + DBC angle

2x = x + DBC angle

DBC angle = x

but in DBC trinagle, DC = BC

Therefore,

DBC angle = CDB angle = x

Therefore in DBC triangle ,

DCB angle = 180 -2x

Now,

DCB angle = 180 -2x

and

ADC angle = 180 -2x (previously proved)

Therefore ADC angle = DCB angle

CDB angle = x (previous proved)

and

DBA angle = x (previously proved)

Therefore DC // AB (Properties of parallelogram)

let ABCD be the cyclicquadrilateral

join AC, AD = DC =>angle DAC = angle DCA

join BD, DC = BC => angle BDC = angle DBC

angle DAC =angle DBC (angels in the same segment)

so angle BDC = angle DCA (from frist two equations)

angle BDA = angle ACB (angles in the same segment)

Add above two

angle (BDC+BDA) = angle (DCA+ACB)

=>angle ADC = angle DCB (proof I)

angle ABC + angle ADC = 180 (opp angles of cyclic quad)

But angle ADC = angle DCB

so angle ABC + angle DCB = 180

so AB//CD (co interior angles are supplementary)

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