In a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P; show that $∠$ APB = $60^{o}$ .

Open in App

Solution

In $\triangle OCD, OD = OC = DC$

$\Rightarrow \triangle OCD$ is equilateral

$\therefore \angle ODC = 60^o$

$\angle ADC + \angle ABC = 180^o$ (ABCD is a cyclic quadrilateral)

$\Rightarrow \angle ODA + 60^o + \angle ABP = 180^o$

$∠ O A D + ∠ A P B = 120^{o}$

$\Rightarrow \angle PAB + \angle ABP = 120^o$

In $\triangle PAB$

$\angle APB = 180^o - \angle PAB - \angle ABP = 180^o-120^o = 60^o$

Suggest Corrections

Similar questions

ABCD is a cyclic quadrilateral of a circle with centre O such that AB is a diameter of this circle and the length of the chord CD is equal to the radius of the circle. If AD and BC produced meet at P, show that $A P B = 60^{\circ} .$

In the following figure, AB is the diameter of a circle with centre O and CD is the chord with length euqal to radius OA.

If AC produced and BD produced meet at point P; show that : $∠ A P B = 60^{\circ} .$

Q. In Fig. 2, a quadrilateral ABCD is drawn to circumscribe a circle, with centre O, in such a way that the sides AB, BC, CD and DA touch the circle at the points, P, Q, R and S respectively. Prove that

A B + C D = B C + D A

Q. In fig. 2, a circle with centre O is inscribed in a quadrilateral ABCD such that, it touches
the sides BC, AB, AD and CD at points P, Q, R and S respectively. If

A B = 29 c m, A D = 23

∠ B = 90^{\circ}

and

D S = 5 c m

, then the radius of the circle (in cm) is:

In the given figure, a circle with centre O, is inscribed in a quadrilateral ABCD such that it touches the sides BC, AB, AD and CD at points P, Q, R and S respectively. If AB = 29 cm, AD = 23 cm, $∠$ = $90^{o}$ and DS = 5 cm then find the radius of the circle.

Join BYJU'S Learning Program

In a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P; show that ∠ APB = 60o.

In is equilateral (ABCD is a cyclic quadrilateral) ∠OAD+∠APB=120o In

In a circle, with centre O, a cyclic quadrilateral ABCD is drawn with AB as a diameter of the circle and CD equal to radius of the circle. If AD and BC produced meet at point P; show that $∠$ APB = $60^{o}$ .