Advertisements
Advertisements
प्रश्न
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
Advertisements
उत्तर
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
संबंधित प्रश्न
In the given figure, ∠PQR = 100°, where P, Q and R are points on a circle with centre O. Find ∠OPR.

ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If ∠DBC = 70°, ∠BAC is 30°, find ∠BCD. Further, if AB = BC, find ∠ECD.
Prove that the line of centres of two intersecting circles subtends equal angles at the two points of intersection.
Bisectors of angles A, B and C of a triangle ABC intersect its circumcircle at D, E and F respectively. Prove that the angles of the triangle DEF are 90°-A, 90° − `1/2 A, 90° − 1/2 B, 90° − 1/2` C.
ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
Circles are described on the sides of a triangle as diameters. Prove that the circles on any two sides intersect each other on the third side (or third side produced).
In the given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =

In the figure, ▢ABCD is a cyclic quadrilateral. If m(arc ABC) = 230°, then find ∠ABC, ∠CDA, ∠CBE.

If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed 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.
