Question

2.ABCD is a rhombus L, M, N are the midpoints of AB, BC and CD respectively. Prove that Z LMN =
90​

Answer 1

$\huge{\text{\underline{Solution:-}}}$

★ Given:-

• ABCD is a rhombus
• P, Q, R are then mid points of AB, BC, CD.

★ To proove that:-

• ∠PQR = 90

★ Construction:-

• Join AC and BD.

★ Proof:-

In ΔDBC

R and Q are the mid points of DC and CB.

∴ RQ // DB $\implies$ ( by mid point theorem)

∴ MQ//ON $\implies$ eq (1)

Now, in Δ ACB

P and Q are the mid points of AB and BC.

∴ AC // PQ $\implies$ (by mid point theorem)

∴ OM //NQ $\implies$ eq (2)

★From eq (1) and (2)

MQ //ON

ON //NQ

Therefore, each pair of opposite side is parallel.

∴ ONQM is a parallelogram

In ONQM

∠MON = 90°

Because diagonals of Rhombus bisect each other at 90°

∠MON = ∠PQR (Opposite angle of parallelogram)

∴ ∠PQR = 90°

Hence proved.

Note:- Here we have assumed ∠LMN as ∠PQR.

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Answer 2

${\bold{\underline{\boxed{Answer:}}}}$

Given : ABCD is a rhombus L, M and N are the mid points of AB, BC, and CD respectively.

To Prove : ∠LMN = 90°

Construction : Join AC and BD

Proof : In the traingle ABC, L and M are the mid-points of the sides AB and BC.

Therefore By the mid-point Theorem

$\implies$ LM || AC ...... (1)

Similarly in the triangle BCD, M and N are the mid-points of the sides BC and CD.

$\implies$ MN || BD ......(2)

Now, from (1) and (2) PM || QR and MR || PQ

Thus PMRQ is a parallelogram.

∠PMR = ∠PQR ( opposite angles of a parallelogram are equal).

Since the diagonals of a rombhus intersect at a right angle.

∠BQC = 90° $\implies$ ∠PQR = 90°

$\implies$ Therefore ∠PMR = 90° $\implies$ ∠LMN = 90°

Hence Proved.

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