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)
APPEARS IN
संबंधित प्रश्न
Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see the given figure). Prove that ∠ACP = ∠QCD.

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.
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.
ABCD is a cyclic quadrilateral in BC || AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
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 cyclic quadrilateral in which AC and BD are its diagonals. If ∠DBC = 55° and ∠BAC = 45°, find ∠BCD.

ABCD is a cyclic quadrilateral in which BA and CD when produced meet in E and EA = ED. Prove that EB = EC.
In the given figure, ABCD is a cyclic quadrilateral in which ∠BAD = 75°, ∠ABD = 58° and ∠ADC = 77°, AC and BD intersect at P. Then, find ∠DPC.

The three angles of a quadrilateral are 100°, 60°, 70°. Find the fourth angle.
