Prove that an isosceles trapezium is always cyclic
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Draw a trapezium ABCD with AB<CD. Drop perpendiculars AM and BN on DC. The solution start as follows:
Given- ABCD is a trapezium with ABIICD and BC=AD(as it is isosceles)
To prove- ABCD is a cyclic quadrilateral
Construction- Drop perpendiculars AM and BN on DC
Proof-In ΔAMD and ΔBNC
AD=BC(given)
angleAMD=angleBNC=90degree
AM=BN (perpedicular distance between two parallel lines is same)
Therefore, ΔAMD CONGRUENT TO ΔBNC (By RHS congruence rule)
angleADC=angleBCD (CPCT) .... (1)
angleBAD+angleADC=180degree (angle on the same side of transversal AD) ....
(2)
FROM (1) and (2)
angleBAD+angleBCD=180degree
⇒Opposite angles are supplementary
Therefore, ABCD is a cyclic quadrilateral.
Hence, proved
Draw a trapezium ABCD with AB<CD. Drop perpendiculars AM and BN on DC. The solution start as follows:
Given- ABCD is a trapezium with ABIICD and BC=AD(as it is isosceles)
To prove- ABCD is a cyclic quadrilateral
Construction- Drop perpendiculars AM and BN on DC
Proof-In ΔAMD and ΔBNC
AD=BC(given)
angleAMD=angleBNC=90degree
AM=BN (perpedicular distance between two parallel lines is same)
Therefore, ΔAMD CONGRUENT TO ΔBNC (By RHS congruence rule)
angleADC=angleBCD (CPCT) .... (1)
angleBAD+angleADC=180degree (angle on the same side of transversal AD) ....
(2)
FROM (1) and (2)
angleBAD+angleBCD=180degree
⇒Opposite angles are supplementary
Therefore, ABCD is a cyclic quadrilateral.
Hence, proved
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Answer:
OPPOSITE ANGLES OF A CYCLIC QUADRILATERAL ARE SUPPLEMENTRY
Hence Proved
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