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
Prove that ‘Opposite angles of a cyclic quadrilateral are supplementary’.
Prove that a cyclic parallelogram is a rectangle.
AC and BD are chords of a circle which bisect each other. Prove that (i) AC and BD are diameters; (ii) ABCD is a rectangle.
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 given figure, O is the centre of the circle such that ∠AOC = 130°, then ∠ABC =

In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to ______.
In the following figure, AOB is a diameter of the circle and C, D, E are any three points on the semi-circle. Find the value of ∠ACD + ∠BED.

If P, Q and R are the mid-points of the sides BC, CA and AB of a triangle and AD is the perpendicular from A on BC, prove that P, Q, R 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.
