if non parallel sides of a trapezium are equal prove that is cyclic
Answers
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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