Question

In the figure, ABCD is a eyelie quadrilateral in which BC = CD and
CF is a tangent to the circle at C. BC is produced to E and
ADCE = 112. if O is the centre of the circle, find (I) angle BOC
(ii) angle dcf​

Answer 1

Solution!!

∠DCB + ∠DCE = 180° (Linear pair)

∠DCB + 112° = 180°

∠DCB = 180° - 112°

∠DCB = 68°

As DC = BC

∴ ∠BCP = ∠DCF

Now, ∠BCP + ∠BCD + ∠BCF = 180°

2∠BCF + 68° = 180°

2∠BCF = 180° - 68°

2∠BCF = 112°

∠BCF = 56°

Also PCF is a tangent and BD is a chord.

∴ ∠BDC = ∠DCF

Arc BD subtends an angle ∠BOC and ∠BDC

∠BOC = 2∠BDC

∠BOC = 2 × 56°

∠BOC = 112°

Some important theorems:-

→ The angle which an arc of a subtends at the centre is double that which it subtends at any point on the remaining part of the circumference.

→ Angles in the same segment of a circle are equal.

→ The angle in a semi-circle is a right angle.

→ The opposite angles of a cyclic quadrilateral (inscribed in a circle) are supplementary.

→ The exterior angle of a cyclic quadrilateral is equal to the interior opposite angle.