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 ∠DCF : ∠F : ∠E = 3 : 5 : 4, Find the angles of the cyclic quadrilateral ABCD.
✓
Answer
Given – In a circle, ABCD is a cyclic quadrilateral AB and DC
Are produced to meet at E and BC and AD are produced to meet at F.
∠DCF : ∠F : ∠E = 3 : 5 : 4
Let ∠DCF = 3X,∠F = 5x,∠E = 4x
Now, we have to find, ∠A,∠B,∠C and ∠D
In cyclic quad. ABCD, BC is produced.
∴ ∠A =∠DCF = 3x
In ΔCDF,
Ext ∠CDA =∠DCF +∠ 3x +5x = 8x
In ΔBCE,
Ext ∠ABC =∠BCE +∠E (∠BCE =∠DCF, Vertically opposite angles)
= ∠DCF +∠E
= 3x + 4x = 7x
Now, in cyclic quad ABCD,
Since, ∠B +∠ = 180°
(since sum of opposite of a cyclic quadrilateral are supplementary)
⇒ 7x + 8x = 180°
⇒15x = 180°
$\Rightarrow x=\frac{180^{\circ}}{15}=12^{\circ}$
$\angle A =3 x =3 \times 12^{\circ}=36^{\circ}$
$\angle B =7 x =7 \times 12^{\circ}=84^{\circ}$
$\angle C =180^{\circ}-\angle A =180^{\circ}-36^{\circ}=144^{\circ}$
$\angle D =8 x =8 \times 12^{\circ}=96^{\circ}$
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