Advertisements
Advertisements
Question
If ABCD is a cyclic quadrilateral in which AD || BC (In the given figure). Prove that ∠B = ∠C.
Advertisements
Solution
It is given that, ABCD is cyclic quadrilateral in which AD || BC
We have to prove `angleB = angle C`
Since, ABCD is a cyclic quadrilateral
So,
`angle B + angle D = 180°` and `angle A + angle C = 180°` ..… (1)
`⇒ angleB + angle A ` = 180° and `angle C + angle D = 180°` (Sum of pair of consecutive interior angles is 180°) …… (2)
From equation (1) and (2) we have
`angleB + angleD + angleB + angle A ` = 360° …… (3)
`angleA + angleC + angleC + angle D ` = 360° …… (4)
`2angleB + angleD + angleA = 2 angleC + angleA + angleD`
`2angleB = 2angleC`
`angleB = angleC`
Hence Proved
APPEARS IN
RELATED QUESTIONS
PA and PB are tangents from P to the circle with centre O. At point M, a tangent is drawn cutting PA at K and PB at N. Prove that KN = AK + BN.
In Fig 2, a circle touches the side DF of ΔEDF at H and touches ED and EF produced at K and M respectively. If EK = 9 cm, then the perimeter of ΔEDF (in cm) is:
In the given figure, if ABC is an equilateral triangle. Find ∠BDC and ∠BEC.
The radius of a circle is 6 cm. The perpendicular distance from the centre of the circle to the chord which is 8 cm in length, is
If O is the centre of a circle of radius r and AB is a chord of the circle at a distance r/2 from O, then ∠BAO =
In following fig. ABC is an equilateral triangle . A circle is drawn with centre A so that ot cuts AB and AC at M and N respectively. Prove that BN = CM.
State, if the following statement is true or false:
Every diameter bisects a circle and each part of the circle so obtained is a semi-circle.
A chord is at a distance of 15 cm from the centre of the circle of radius 25 cm. The length of the chord is
If a hexagon ABCDEF circumscribe a circle, prove that AB + CD + EF = BC + DE + FA.
Is every chord of a circle also a diameter?
