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प्रश्न
In the given figure, AB = AC. Prove that DECB is an isosceles trapezium.
उत्तर
Here, AB = AC
`=>` ∠B = ∠C
∴ DECB is a cyclic quadrilateral
(In a triangle, angles opposite to equal sides are equal)
Also, ∠B + ∠DEC = 180° ...(1)
(Pair of opposite angles in a cyclic quadrilateral are supplementary)
`=>` ∠C + ∠DEC = 180° ...[From (1)]
But this is the sum of interior angles
On one side of a transversal.
∴ DE || BC
But ∠ADE = ∠AED = ∠C ...[Corresponding angles]
Thus, ∠ADE = ∠AED
`=>` AD = AE
`=>` AB – AD = AC – AE ...(∴ AB = AC)
`=>` BD = CE
Thus, we have, DE || BC and BD = CE
Hence, DECB is an isosceles trapezium
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संबंधित प्रश्न
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In the figure, ABCD is a cyclic quadrilateral with BC = CD. TC is tangent to the circle at point C and DC is produced to point G. If angle BCG=108° and O is the centre of the circle, find: angle DOC
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Prove that the quadrilateral formed by angle bisectors of a cyclic quadrilateral ABCD is also cyclic.
In the adjoining figure, AB is the diameter of the circle with centre O. If ∠BCD = 120°, calculate:
(i) ∠BAD (ii) ∠DBA
In the given circle with centre O, ∠ABC = 100°, ∠ACD = 40° and CT is a tangent to the circle at C. Find ∠ADC and ∠DCT.
An exterior angle of a cyclic quadrilateral is congruent to the angle opposite to its adjacent interior angle, to prove the theorem complete the activity.
Given: ABCD is cyclic,
`square` is the exterior angle of ABCD
To prove: ∠DCE ≅ ∠BAD
Proof: `square` + ∠BCD = `square` .....[Angles in linear pair] (I)
ABCD is a cyclic.
`square` + ∠BAD = `square` ......[Theorem of cyclic quadrilateral] (II)
By (I) and (II)
∠DCE + ∠BCD = `square` + ∠BAD
∠DCE ≅ ∠BAD
