Question

If non-parallel sides of a trapezium are

equal, prove that it is cyclic.​

Answer 1

$\large\underline{\sf{Solution-}}$

Let ABCD be trapezium such that

• AB || CD

and

• AD = BC.

NOTE :- To show that trapezium is cyclic, its enough to show that sum of the opposite angles is supplementary.

Construction :- Construct AL and BM perpendicular on CD meeting CD at L and M respectively.

Now,

$\rm :\longmapsto\:In \: \triangle \: ALD \: and \: \triangle \: BMC$

$\rm :\longmapsto\:AD \: = \: BC \: \: \{given \}$

$\rm :\longmapsto\: \angle \: ALD \: = \angle \: BMC \: \: \{each \: 90 \degree \}$

$\rm :\longmapsto\:AL \: = \: BM \: \: \{distance \: between \parallel \: lines \}$

$\bf\implies \:\triangle \: ALD \: \cong \: \triangle \: BMC \: \: \{RHS \: Congruency \}$

$\rm :\implies\: \angle \: ADL \: = \angle \: BCM \: \: \{CPCT \}$

$\rm :\implies\: \angle \: ADC \: = \angle \: BCD \: - - - (1)$

Now,

• As ABCD is a trapezium and AB || CD

$\rm :\implies\:\angle \: DAB \: + \: \angle \: ADC = 180 \degree \: \{sum \: of \: cointeriors \}$

$\bf\implies \:\angle \: DAB \: + \: \angle \: BCD \: = \: 180 \degree \: \: \{using \: (1) \}$

$\bf\implies \:ABCD \: is \: cyclic \: trapezium.$

Additional Information :-

• All the four vertices of a quadrilateral inscribed in a circle lie on the circumference of the circle.

• The sum of two opposite angles in a cyclic quadrilateral is equal to 180 degrees (supplementary angles)

• The measure of an exterior angle is equal to the measure of the opposite interior angle.

• The product of the diagonals of a quadrilateral inscribed in a circle is equal to the sum of the product of its two pairs of opposite sides.

• The perpendicular bisectors of the four sides of the inscribed quadrilateral intersect at the center O.