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In a quadrilateral klmn ,kn=lm and angles knm and lmn are equal. Prove that the points k,l,m and n lie on a circle.

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8 months ago

```							9Given: KLMN is a quadrilateral. KN=LM, angle KNM = angle LMN. To prove: That the quadrilateral is cyclic, i.e All points lie on the circle. Proof: Side KN = Side LM.......................{Given} Therefore, opposite sides of the quadrilateral are equal. Therefore, the quadrilateral KLMN is a Rectangle. Now, A quadrilateral is cyclic only when their opposite angles form 180° But all angles of a rectangle are 90° Therefore, Angle KNM + Angle KLM 90° + 90° 180° Therefore, quadrilateral KLMN is cyclic. Therefore points K, L, M, N lie on the circle.
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8 months ago
```							dear student, note that aruns answer is totally rubbish, misleading and WRONG.draw the fig KLMN. join KM and LN.now, angle KNM= angle LMN (given)NM= MN (common side)KN= LM (given)by SAS, tri KNM is congruent to tri LMN.by CPCT, KM= LN......(1)now, KL= LK (common side)KN= LM (given)LN= KM (from 1)hence by SSS, tri KNL is congruent to tri LMK.by CPCT, angle NKL= angle MLK= x (say).also, given angle KNM= angle LMN= y (say)so that x+x+y+y= 360or x+y= 180or angle MLK + angle KNM= 180.since the sum of opposite angles of a quad being 180 deg implies its cyclic nature (converse of cyclic quad theorem), hence we deduce that the points k,l,m and n lie on a circle.further, note that such a quad in general shall be a trapezium, not a rectangle as arun mentioned.KINDLY APPROVE :))
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8 months ago
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