#### 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.

#### Solution

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°

⇒ `x = (180°)/15 = 12° `

∠A = 3x = 3 × 12° = 36°

∠B = 7x = 7×12° = 84°

∠C = 180° - ∠A = 180° - 36° = 144°

∠D = 8x = 8 × 12° = 96°