Let $□$ MNOP be a rhombus. Q is a point in the interior of rhombus MNOP. Such that QM = QO. Prove that Q lies on diagonal NP.

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Solution

Consider $△$ PQM and $△$ PQO,
PM = PO (MNOP is a rhombus)
PQ = PQ (Common)
QM = QO (Given)
By SSS congruency, $△$ PQM $≅$ $△$ PQO.
$∠$ PQM = $∠$ PQO (by c.a.c.t) ...(1)

Now, consider $△$ NQM and $△$ NQO.
NM = NO (MNOP is a rhombus)
NQ = NQ (Common)
QM = QO (Given)
By SSS congruency, $△$ NQM $≅$ $△$ NQO.
$∠$ NQM = $∠$ NQO (by c.a.c.t.) ...(2)

Now,
$∠$ PQM + $∠$ PQO + $∠$ NQM + $∠$ NQO= 360 $°$ (Q is a point)
Or, 2( $∠$ PQO + $∠$ NQO) = 360 $°$ [From equations (1) & (2)]
Or, $∠$ PQO + $∠$ NQO = 180 $°$
Hence, PN is a straight line on which point Q lies.

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Let □MNOP be a rhombus. Q is a point in the interior of rhombus MNOP. Such that QM = QO. Prove that Q lies on diagonal NP.

Let $□$ MNOP be a rhombus. Q is a point in the interior of rhombus MNOP. Such that QM = QO. Prove that Q lies on diagonal NP.