In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal. Prove it.
A cyclic trapezium ABCD in which AB ∥ DC and AC and BD are joined.
(i) AD = BC
(ii) AC = BD
∵ chord AD subtends ∠ABD and chord BC subtends BDC
At the circumference of the circle.
But ∠ABD = ∠BDC [proved]
Chord AD = Chord BC
⇒ AD = BC
Now in ∆ADC and ∆ BCD
DC = DC [Common]
∠CAD = ∠CBD [angles in the same segment]
And AD = BC [proved]
By Side – Angle – Side criterion of congruence, we have
∴ ∆ADC ≅ ∆BCD [ SAS axion]
The corresponding parts of the congruent triangle are congruent
∴ AC = BD [c.p.c.t]
Video Tutorials For All Subjects
- Arc and Chord Properties - Angles in the Same Segment of a Circle Are Equal (Without Proof)