Answer. Question :- Prove that in a cyclic trapezium the non parallel sides are equal. Answer :- Let ABCD be the cyclic trapezium in which AB||CD. Also chord AD subtends ∠ABD and chord BC subtends ∠BDC on the circle at B and D respectively.
In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.
In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.Let ABCD be the cyclic trapezium in which AB∣∣DC, AC and BD are the diagonals.
Required to prove:
(i) AD=B
(ii) AC=BD
Proof:
It’s seen that chord AD subtends ∠ABD and chord BC subtends ∠BDC at the circumference of the circle.
But, ∠ABD=∠BDC [Alternate angles, as AB∣∣DC with BD as the transversal]
Answers & Comments
Answer:
Answer. Question :- Prove that in a cyclic trapezium the non parallel sides are equal. Answer :- Let ABCD be the cyclic trapezium in which AB||CD. Also chord AD subtends ∠ABD and chord BC subtends ∠BDC on the circle at B and D respectively.
mark it Braillent friend
Answer:- Here,
Hey mate,
In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.
In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal.Let ABCD be the cyclic trapezium in which AB∣∣DC, AC and BD are the diagonals.
Required to prove:
(i) AD=B
(ii) AC=BD
Proof:
AD=BC
DC=DC [Common]
∠CAD=∠CBD [Angles in the same segment are equal]
AD=BC [Proved above]
ΔADC≅ΔBCD
Therefore, by CPCT
AC=BD
Good day,
Thanks:)