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Question
In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal. Prove it.
Solution
A cyclic trapezium ABCD in which AB || DC and AC and BD are joined.
To prove:
- AD = BC
- 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 ∆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 axiom]
The corresponding parts of the congruent triangle are congruent
Therefore, AC = BD ...[c.p.c.t]
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