Advertisements
Advertisements
Question
ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that AD || BC .
Advertisements
Solution
If ABCD is a cyclic quadrilateral in which AB and CD when produced meet in E such that EA = ED, then we have to prove the following, AD || BC
It is given that EA = ED, so
Since, ABCD is cyclic quadrilateral
`x + angleABC = 180 ⇒ angleDAB = 180 - x`
And ; ` x + angleBCD = 180 ⇒ angle BCD = 180- x `
Now,
`angle DAB + angle ABC = x + 180 - x = 180`
Therefore, the adjacent angles `angleDAB ` and `angleABC` are supplementary
Hence, AD || BC
APPEARS IN
RELATED QUESTIONS
A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
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.
The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at distance 4 cm from the centre, what is the distance of the other chord from the centre?
In the given figure, ∠BAD = 78°, ∠DCF = x° and ∠DEF = y°. Find the values of x and y.

In a cyclic quadrilateral ABCD, if ∠A − ∠C = 60°, prove that the smaller of two is 60°
Prove that the circles described on the four sides of a rhombus as diameters, pass through the point of intersection of its diagonals.
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.

In a cyclic quadrilaterals ABCD, ∠A = 4x, ∠C = 2x the value of x is
ABCD is a cyclic quadrilateral such that ∠A = 90°, ∠B = 70°, ∠C = 95° and ∠D = 105°.
