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In triangle PQR seg XY || seg QR , M and N are midpoints of seg PY and Seg PR then prove that 1)∆PQR ~∆PQN. .2)Seg XM || seg QN

User BCA
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Explanation:

1)∆PQR ~∆PQN. .2)Seg XM || seg QN

To prove that ∆PQR ~ ∆PQN, we need to show that their corresponding sides are proportional in length.

Using the midpoint theorem, we know that PM = MY and PN = NR.

Therefore, we can write:

QR = QP + PR

QP = QR - PR

and

QN = QP/2

PR = QR/2

PN = NR - PR = QR/2 - QR/2 = 0

Using these expressions, we can write:

QP/QR = (QR - PR)/QR = 1 - PR/QR = 1 - 1/2 = 1/2

and

PN/PQ = 0/QP = 0

Since two sides of ∆PQR and ∆PQN are proportional, we can conclude that ∆PQR ~ ∆PQN.

To prove that seg XM || seg QN, we can use the fact that M and N are midpoints of PY and PR, respectively.

Since seg XY || seg QR, we know that ∠QXY = ∠QRP (corresponding angles).

Also, since M is the midpoint of PY, we know that seg PM is parallel to seg QY (midpoint theorem).

Therefore, we have:

∠NQM = ∠QXY (alternate angles)

∠NQM = ∠QRP (corresponding angles)

∠PMX = ∠QYX (alternate angles)

∠PMX = ∠QRP (corresponding angles)

Since two pairs of alternate angles are equal, we can conclude that seg XM || seg QN (by the converse of the alternate interior angles theorem).

User Shirish Coolkarni
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