Question

If two non parallel sides of a trapezium are equal prove tjat it is cyclic

Answer 1

Solution⤵️

Let ABCD be a trapezium
in which AB=DC

To prove: ABCD is a cyclic quadrilateral i.e.,<4+<1=180°

DE parallel to AB is drawn
Now, AD is parallel to BE (since AD||BC)
=>ABED is a parallelogram
=>AB=DE (opposite sides of a parallelogram)
=> DE=DC (since AB=DC)
=>∆DEC is a isosceles triangle
=> <1=<2 ➡️[1] (angles opp. to equal sides are equal)

also, <4=<3 ➡️[2] (opp. angles of parallelogram ABED are equal)

Now,
<3+<2=180° (linear pair)
=><4+<1=180° 【from[1)]&[2]】
thus, ABCD is a cyclic quadrilateral

hence proved

Answer 2

Hello mate =_=

____________________________

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______❤

_____________________________❤