Two circles intersect at P and Q. Through P, two straight lines APB and CPD are drawn
to meet the circles at A, B, C and D. AC and DB when produced meet at O. Show that
OAQB is a cyclic quadrilateral.
dagram
Answers
OAQB is a cyclic Quadrilateral
Step-by-step explanation:
ACPQ is cyclic Quadrilateral
=> ∠ACP + ∠AQP = 180° ( Sum of opposite angles of cyclic quadrilateral)
∠ACP = ∠ACD
∠ACD = ∠COD + ∠ODP
=> ∠ACP = ∠AOB + ∠BDP
∠BDP = ∠BQP ( angle by same chord BP )
=> ∠ACP = ∠AOB + ∠BQP
∠ACP + ∠AQP = 180°
=> ∠AOB + ∠BQP + ∠AQP = 180°
∠BQP + ∠AQP = ∠AQB
=> ∠AOB + ∠AQB = 180°
=> OAQB is a cyclic Quadrilateral
QED
Proved
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Proved that OAQB is a cyclic quadrilateral.
To show : OAQB is a cyclic quadrilateral
Given :
- Two circles intersect at P and Q.
- Through P, two straight lines APB and CPD are drawn.
- It meets the circles at A, B, C and D.
- AC and DB when produced meet at O.
Quadrilateral OAQB will be con-cyclic when : ∠O + ∠AQB = 180°.
At first join PQ.
From the figure :
∠PQB = ∠PDB -----> ( 1 )
∠ACP + ∠AQP = 180° -----> ( 2 )
Sum of interior opposite angles.
∠ACP = ∠ACD = ∠O + ∠CDB
From equation ( 2 ), substitute ∠ACP we get
180° - ∠AQP = ∠O + ∠CDB
From equation ( 1 ), we get
180° = ∠AQP + ∠O + ∠PQB
180° = ∠O + ∠AQP + ∠PQB = ∠O + ∠AQB
Hence proved that, OAQB is a cyclic quadrilateral.
To learn more...
1. Two circles intersect at P and Q Through P two straight lines APB and CQD are drawn to meet the circles at AB.C and D AC and BD when produced meet at O Show that OAQB is a cyclic quadrilateral
brainly.in/question/14416309
2. In the following diagrams, ABCD is a square and APB is an equilateral triangle .in each case, prove that triangle APD is congruent triangle BPC and find the angles of triangle DPC.
brainly.in/question/737448
