Advertisements
Advertisements
Question
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
Advertisements
Solution
It is given that ∠A – ∠C = 60° and ABCD is a cyclic quadrilateral.

We have to prove that smaller of two is 60°
Since ABCD is a cyclic quadrilateral
So ∠A + ∠C = 180° (Sum of opposite pair of angles of cyclic quadrilateral is 180°) ..… (1)
And,
∠A – ∠C = 60° (Given) ..… (2)
Adding equation (1) and (2) we have
`2angle A = 240° `
`angle A =( 240° )/2`
= 120°
So, ∠C = 60°
Hence, smaller of two is 60°.
APPEARS IN
RELATED QUESTIONS
ABC and ADC are two right triangles with common hypotenuse AC. Prove that ∠CAD = ∠CBD.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Two congruent circles intersect each other at points A and B. Through A any line segment PAQ is drawn so that P, Q lie on the two circles. Prove that BP = BQ.

In the figure m(arc LN) = 110°,
m(arc PQ) = 50° then complete the following activity to find ∠LMN.
∠ LMN = `1/2` [m(arc LN) - _______]
∴ ∠ LMN = `1/2` [_________ - 50°]
∴ ∠ LMN = `1/2` × _________
∴ ∠ LMN = __________
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
In the figure, ▢ABCD is a cyclic quadrilateral. If m(arc ABC) = 230°, then find ∠ABC, ∠CDA, ∠CBE.

ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
ABCD is a parallelogram. A circle through A, B is so drawn that it intersects AD at P and BC at Q. Prove that P, Q, C and D are concyclic.
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.
