Question

prove that if two non parallel sides of a Trapezium are equal then is it cyclic​

Answer 1

Given: :- ABCD is a trapezium where AB||CD and AD = BC

To prove: ABCD is cyclic.

Construction: Draw DL⊥AB and CM⊥AB.

Proof: In ΔALD and ΔBMC,

AD = BC (given)

DL = CM (distance between parallel sides)

∠ALD = ∠BMC (90°)

ΔALD ≅ ΔBMC (RHS congruence criterion)

⇒ ∠DAL = ∠CBM (C.P.C.T) (1)

Since AB||CD,

∠DAL + ∠ADC = 180° (sum of adjacent

interior angles is supplementary)

⇒ ∠CBM + ∠ADC = 180° (from (1))

⇒ ABCD is a cyclic trapezium (Sum of opposite angles is supplementary)

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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