Question

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

Answer 1

GIVEN : non parallel sides AD and BC of a trapezium are epual
TO PROVE : ABCD is a cyclic trapezium
Construction : draw DE perpendicular to AB and CF perpendicular to AB
Proof
in triangle DEA and CFB
AD equal to BC
Angle DEA equal to Angle CFA
DE equal to CF
By RHS
triangle DEA is congruent to triangle CFA
Angle A is equal to angle B
Angle ADE is equal to angle BCF
90 degree plus angle ADE is equal to 90 degree plus angle BCF
angle EDC plus angle ADE is equal to angle FCD plus angle BCF
angle ADC is equal to angle BCD
angle D is equal to angle C
angle A is equal to angle B and angle C is equal to angle D
angle A plus angle B plus angle C plus angle D is equal to 360 degree
two angle B plus two angle D is equal to 360 degree
angle B plus angle D is equal to 180 degree
Hence, ABCD is a cyclic trapezium.

Answer 2

Hello mate =_=

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Solution:

It is given that ABCD is a trapezium with AB∥CD and AD=BC

We need to prove that ABCD is a cyclic quadrilateral.

Construction: Draw AM⊥CD and BN⊥CD

In ∆AMD and ∆BNC, we have

AD=BC            (Given)

∠AMD=∠BNC          (Each equal to 90°)

AM=BN        (Distance between two parallel lines is constant.)

Therefore, by RHS congruence rule, we have ∆AMD≅∆BNC

⇒∠D=∠C        (Corresponding parts of congruent triangles are equal)   ........ (1)

We also have ∠A+∠D=180′      (Co-interior angles, AB∥CD)     ......... (2)

From (1) and (2), we can say that ∠A+∠C=180°

⇒ ABCD is a cyclic quadrilateral.

(If the sum of a pair of opposite angles of a quadrilateral is 180°, the quadrilateral is cyclic.)

I hope, this will help you.

Thank you______❤

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