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
If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
Let the vertex of an angle ABC be located outside a circle and let the sides of the angle intersect equal chords AD and CE with the circle. Prove that ∠ABC is equal to half the difference of the angles subtended by the chords AC and DE at the centre.
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
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.
In the given figure, ABCD is a cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
ABCD is a cyclic quadrilateral. M (arc ABC) = 230°. Find ∠ABC, ∠CDA, and ∠CBE.
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
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.
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.
