Question

please answer fast as possible​

Answer 1

To Prove : angle ADC + angle BCD = 180

Proof : Draw seg PQ

PECQ is a cyclic quadrilateral and angle PQD

angle PQD = angle PBC ...... ( 1 )

........[Corollary of cyclic quadrilateral] theoram)

PQDA is a cyclic quadrilateral

angle PQD + angle PMD = 180

.....[Theoram on cyclic quadrilateral]

angle PBC + angle PAD = 180 ....[from 1]

angle ADC + angle BCD =180 ....[B-P-A]

.....[Proved]

also,

seg SQ || seg RP

.....[ Interior angles test for parallel lines]

Answer 2

Question: Two circles intersect at a point P and Q. Secants through P and Q intersect the circle at AB and DC. Prove that: ∠ADC + ∠BCD = 180°

Proof:

Join P and Q, to form a line PQ.

APQD is a cyclic quadrilateral.

∴ ∠ADC + ∠APQ = 180° [Eq(1)]

[Opposite angles in a cyclic quadrilateral are supplementary]

BPQC is a cyclic quadrilateral.

∴ ∠BCD + ∠BPQ = 180° [Eq(2)]

[Opposite angles in a cyclic quadrilateral are supplementary]

∠APQ + ∠BPQ = 180° [Linear Pair]

Let the above be Eq(3)

Adding Equation 1 and 2 we get,

∠ADC + ∠APQ + ∠BCD + ∠BPQ = 180° + 180°

Substitute Eq(3) above.

∠ADC + ∠BCD + 180° = 360°

∠ADC + ∠BCD = 360° - 180°

∠ADC + ∠BCD = 180°

Hence Proved.