A cyclic trapezium ABCD in which AB ∥ DC and AC and BD are joined.
To prove-
(i) AD = BC
(ii) AC = BD
Proof:
∵ 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 ΔADC and Δ BCD
DC = DC [Common]
∠CAD = ∠CBD [angles in the same segment]
And AD = BC [proved]
By Side – Angle – Side criterion of congruence, we have
ΔADC ≅ ΔBCD [ SAS axion]
The corresponding parts of the congruent triangle are congruent
Therefore, AC = BD [c.p.c.t]
