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prove that a cyclic trapezium is always isosceles trapezium.

Prove that the line segment joining the mid-points of opposites sides of a quadrilateral bisect each other.

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Prove that a cyclic trapezium is always isosceles trapezium.

In a cyclic trapezium,

 

∠BAD + ∠BCD = 180°

∠BAD + ∠ABC = 180°

 

∴ ∠ABC = ∠BCD

 

In ∆ABC and ∆BDC,

 

⇒∠ABC = ∠BCD

⇒∠BAC = ∠CDB (angles in the same segment)

⇒BC = BC (common side)

⇒∆ABC ≅ ∆BDC

 

∴ AB = CD and AC = BD

 

Hence, cyclic trapezium ABCD is isosceles and diagonals are equal to each other.

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Let ABCD is a quadrilateral in which P, Q, R, and S are the mid-points of sides AB, BC, CD, and DA respectively. Join PQ, QR, RS, SP, and BD.

In ΔABD, S and P are the mid-points of AD and AB respectively. Therefore, by using mid-point theorem, it can be said that

SP || BD and SP =   BD ... (1)

Similarly in ΔBCD,

QR || BD and QR =  BD ... (2)

From equations (1) and (2), we obtain

SP || QR and SP = QR

In quadrilateral SPQR, one pair of opposite sides is equal and parallel to

each other. Therefore, SPQR is a parallelogram.

We know that diagonals of a parallelogram bisect each other.

Hence, PR and QS bisect each other.

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In a cyclic trapezium,

∠BAD + ∠BCD = 180°

∠BAD + ∠ABC = 180°

∴ ∠ABC = ∠BCD

In ∆ABC and ∆BDC,

∠ABC = ∠BCD

∠BAC = ∠CDB (angles in the same segment)

BC = BC (common side)

∴ ∆ABC ≅ ∆BDC

∴ AB = CD and AC = BD

Hence, cyclic trapezium ABCD is isosceles and diagonals are equal to each other.

 
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