In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal. Prove it. - Noon Academy

In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal. Prove it.

Circles | Class 10 | Exercise 1 | Exercise 17B | ICSE | Maths | Selina

In a cyclic-trapezium, the non-parallel sides are equal and the diagonals are also equal. Prove it.

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(B) - 1

Solution:

Let

$ABCD$

be the cyclic trapezium in which

$AB\text{ }||\text{ }DC,\text{ }AC\text{ }and\text{ }BD$

are the diagonals.

Required to prove:

$\left( i \right)\text{ }AD\text{ }=\text{ }BC$

$\left( ii \right)\text{ }AC\text{ }=\text{ }BD$

Proof:

It’s seen that

$chord\text{ }AD$

subtends

$\angle ABD\text{ }and\text{ }chord\text{ }BC$

subtends

$\angle BDC$

at the circumference of the circle.

But,

$\angle ABD\text{ }=\angle BDC$

[Alternate angles, as AB || DC with BD as the transversal]

So,

$Chord\text{ }AD$

must be equal to

$chord\text{ }BC$

$AD\text{ }=\text{ }BC$

Now, in

$\vartriangle ADC\text{ }and\text{ }\vartriangle BCD$

$DC\text{ }=\text{ }DC$

[Common]

$\angle CAD\text{ }=\angle CBD$

[Angles in the same segment are equal]

$AD\text{ }=\text{ }BC$

[Proved above]

By SAS criterion of congruence

$\vartriangle ADC\cong \vartriangle BCD$

By CPCT

$AC\text{ }=\text{ }BD$

i

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