Advertisements
Advertisements
Question
If bisectors of opposite angles of a cyclic quadrilateral ABCD intersect the circle, circumscribing it at the points P and Q, prove that PQ is a diameter of the circle.
Advertisements
Solution
Given: ABCD is a cyclic quadrilateral.
DP and QB are the bisectors of ∠D and ∠B, respectively.
To prove: PQ is the diameter of a circle.
Construction: Join QD and QC.
Proof: Since, ABCD is a cyclic quadrilateral.
∴ ∠CDA + ∠CBA = 180° ...[Sum of opposite angles of cyclic quadrilateral is 180°]
On dividing both sides by 2, we get
`1/2 ∠CDA + 1/2 ∠CBA = 1/2 xx 180^circ = 90^circ`
⇒ ∠1 + ∠2 = 90° ...(i) `[∠1 = 1/2 ∠CDA "and" ∠2 = 1/2 ∠CBA]`
But ∠2 = ∠3 [Angles in the same segment QC are equal] ...(ii)
∠1 + ∠3 = 90°
From equations (i) and (ii),
∠PDQ = 90°
Hence, PQ is a diameter of a circle, because diameter of the circle.
Subtends a right angle at the circumference.
APPEARS IN
RELATED QUESTIONS
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.
If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
ABCD is a cyclic quadrilateral in ∠BCD = 100° and ∠ABD = 70° find ∠ADB.
ABCD is a cyclic trapezium with AD || BC. If ∠B = 70°, determine other three angles of the trapezium.
In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.
ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.
Find all the angles of the given cyclic quadrilateral ABCD in the figure.
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
