Advertisements
Advertisements
प्रश्न
ABCD is a parallelogram. The circle through A, B and C intersect CD (produced if necessary) at E. Prove that AE = AD.
Advertisements
उत्तर

It can be observed that ABCE is a cyclic quadrilateral and in a cyclic quadrilateral, the sum of the opposite angles is 180°.
∠AEC + ∠CBA = 180°
∠AEC + ∠AED = 180° (Linear pair)
∠AED = ∠CBA ... (1)
For a parallelogram, opposite angles are equal.
∠ADE = ∠CBA ... (2)
From (1) and (2),
∠AED = ∠ADE
AD = AE (Angles opposite to equal sides of a triangle)
संबंधित प्रश्न
Prove that the circle drawn with any side of a rhombus as diameter passes through the point of intersection of its diagonals.
In the given figure, ABCD is a cyclic quadrilateral. Find the value of x.

ABCD is a cyclic quadrilateral in ∠DBC = 80° and ∠BAC = 40°. Find ∠BCD.
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).
In the given figure, ABCD is a quadrilateral inscribed in a circle with centre O. CD is produced to E such that ∠AED = 95° and ∠OBA = 30°. Find ∠OAC.

PQRS is a cyclic quadrilateral such that PR is a diameter of the circle. If ∠QPR = 67° and ∠SPR = 72°, then ∠QRS =
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°.
If a line is drawn parallel to the base of an isosceles triangle to intersect its equal sides, prove that the quadrilateral so formed is cyclic.
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.

