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Sum of Opposite Angles of a Cyclic Quadrilateral

ABCD is a cyc...

Question

ABCD is a cyclic quadrilateral. Sides AB and DC produced meet at point E; whereas sides BC and AD produced meet at point F.

If $∠ D C F : ∠ F : ∠ E = 3 : 5 : 4,$ find the angles of the cyclic quadrilateral ABCD.

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Solution

Given $DCF : F : E = 3:4:5$

In $\triangle CDF: \angle ADC = 3x+5x = 8x$

In $\triangle BCE: \angle ABC = 3x+4x = 7x$

$\angle ABC + \angle ADC = 180^o$

or $7x+8x= 180 \Rightarrow x = 12^o$

Therefore

$\angle DAC = 180 -12x = 180 - 144 = 36^o$

$\angle CBA = 7x = 7 \times 12 = 84^o$

$\angle DCB = 180-3x = 180-36 = 144^o$

$\angle ADC = 8x = 8 \times 12 = 96^o$

$\\$

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