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ABCD is a cyclic quadrilateral - Leaving Cert Mathematics - Question 6B - 2011

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Question 6B

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ABCD is a cyclic quadrilateral. The opposite sides, when extended, meet at P and Q, as shown. The angles \( \alpha, \beta \) and \( \gamma \) are as shown. Prove tha... show full transcript

Worked Solution & Example Answer:ABCD is a cyclic quadrilateral - Leaving Cert Mathematics - Question 6B - 2011

Step 1

Show that angles in a cyclic quadrilateral are supplementary

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Answer

In cyclic quadrilateral ABCD, the opposite angles sum to 180 degrees. Therefore, ( \angle A + \angle C = 180^ ext{\circ} ) and ( \angle B + \angle D = 180^ ext{\circ} ). This property is fundamental to cyclic quadrilaterals.

Step 2

Determine angles at points P and Q

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Answer

At point P, ( \angle APB = \angle A + \angle B ), and at point Q, ( \angle DQC = \angle D + \angle C ). Since ( A + C = 180^ ext{\circ} ) and ( B + D = 180^ ext{\circ} ), we also have ( \angle APB + \angle DQC = 360^ ext{\circ} ).

Step 3

Find relationships involving angles \( \beta \) and \( \gamma \)

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Answer

From the cyclic properties, we know ( \angle D = \beta ) and ( \angle C = \gamma ). Therefore, ( \beta + \gamma = 180^ ext{\circ} - \angle A - \angle B ). Using the fact that ( A + B = 180^ ext{\circ} - 2\alpha ) (as derived from the properties of the angles at points P and Q), we can combine these results.

Step 4

Conclude the proof

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Answer

Substituting ( A + B = 180^ ext{\circ} - 2\alpha ) into ( \beta + \gamma = 180^ ext{\circ} - (A + B) ), we have: [ \beta + \gamma = 180^ ext{\circ} - (180^ ext{\circ} - 2\alpha) = 2\alpha. ] This concludes the proof that ( \beta + \gamma = 180^ ext{\circ} - 2\alpha . )

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