Question

Guys, please answer this above question. Either with picture solution or typed solution​

Answer 1

Required Answer:-

Given:

• The non parallel sides of a trapezium are equal.

To prove:

• The given quadrilateral is cyclic.

Proof:

Let us assume that, ABCD is the trapezium where AB || DC and non parallel sides are equal i.e, AD = BC.

Construction: Draw DE ⊥ AB and CF ⊥ AB

➡ To prove that it is a cyclic quadrilateral, we have to prove that sum of one pair of opposite angles is 180°

Now, observe the image carefully.

In ΔADE and ΔBCF,

1. ∠AED = ∠BFC (90° each as AE ⊥ DC and BF ⊥ DC)
2. AD = BC (Given)
3. DE = CF (Distance between two parallel sides is equal)

∴ ΔADE ≅ ΔBCF (R.H.S)

Thus,

∠DAE = ∠CBF (C.P.C.T)

➡ ∠A = ∠B ......(i)

Now, for parallel lines AB and CD,

➡ AD is the transversal line.

➡ ∠A + ∠D = 180°

➡ ∠B + ∠D = 180° (From (i))

So, in ABCD, sum of one pair of opposite angles is 180°.

So, ABCD is a cyclic quadrilateral. (Hence Proved)

Answer 2

Answer:

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