Question

if the non-parallel of a trapezium are equal, prove that it is cyclic​

Answer 1

Given:

ABCD is a trapezium where non-parallel sides AD and BC are equal.

Construction:

DM and CN are perpendicular drawn on AB from D and C, respectively.

To prove:

ABCD is cyclic trapezium.

Proof:

In △DAM and △CBN,

AD=BC       ...[Given]

∠AMD=∠BNC      ...[Right angles]

DM=CN     ...[Distance between the parallel lines]

Therefore, △DAM≅△CBN  by RHS congruence condition.

Now, ∠A=∠B      ...[by CPCT]

Also, ∠B+∠C=180°    ....[Sum of the co-interior angles]

⇒∠A+∠C=180°.

Answer 2

Appropriate Question :-

If the non-parallel sides of a trapezium are equal, prove that it is cyclic.

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

Let assume that ABCD be a trapezium with AB || CD and AD = BC.

Now, we have to prove that ABCD is a cyclic trapezium. So, it is enough to show that sum of the either pair of opposite angles are supplementary.

Construction :- From A and B, draw AE and BF perpendicular on CD intersecting CD at E and F respectively.

Now, In right triangle AED and BFC

$\rm \: AE \: = \: BF \: \: \{distance \: between \: parallel \: lines \}$

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

$\rm \: \angle \: AED \: = \: \angle BFC \: \: \: \{ \: each \: 90 \degree \: \}$

$\rm\implies \: \triangle \: AED \: = \: \triangle BFC \: \: \: \{ \: RHS \: Rule \: \}$

$\rm\implies \:\angle 1 = \angle 2 \: \{CPCT \}$

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

Now as it is given that

$\rm \: AB \: \parallel \: CD$

$\rm\implies \:\angle ABC + \angle BCD = 180 \degree \: \{sum \: of \: cointerior \}$

Now, using equation (1), we get

$\rm\implies \:\angle ABC + \angle ADC = 180 \degree \:$

$\rm\implies \: ABCD \: is \: a \: cyclic \: trapezium \:$

$\rule{190pt}{2pt}$

Additional Information :-

1. Angle is semi-circle is right angle.

2. Angle in same segments are equal.

3. Angle subtended at the centre of a circle by an arc is double the angle subtended on the circumference by the same arc.

4. Perpendicular drawn from centre bisects the chord.