Advertisements
Advertisements
Question
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.
Advertisements
Solution
Since, A, C, D and E are four point on a circle, then ACDE is a cyclic quadrilateral.
∠ACD + ∠AED = 180° ...(i) [Sum of opposite angles in a cyclic quadrilateral is 180°]
Now, ∠AEB = 90° ...(ii)
We know that, diameter subtends a right angle to the circle.
On adding equations (i) and (ii), we get
(∠ACD + ∠AED) + ∠AEB = 180° + 90° = 270°
⇒ ∠ACD + ∠BED = 270°
APPEARS IN
RELATED QUESTIONS
Prove that a cyclic parallelogram 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.
Two chords AB and CD of lengths 5 cm 11cm respectively of a circle are parallel to each other and are on opposite sides of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle.
In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
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).
Prove that the centre of the circle circumscribing the cyclic rectangle ABCD is the point of intersection of its diagonals.
Find all the angles of the given cyclic quadrilateral ABCD in the figure.
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
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.
