In the figure ABCD is a parallelogram. P, Q, R and S are the midpoint of the sides of the parallelogram. Prove that PQ = RS and QR = PS.

ΔRCQ = ΔAPS

QR = PS (CPCTC)

ΔSRD ≅ ΔQPB

PQ = RS (CPCTC)

Step-by-step explanation:

Given that ABCD is a parallelogram, we have;

AD = BC, and DC = AB (Definition of a parallelogram)

AD = AS + SD Given S is midpoint of AD

Similarly,

DC = DR + RC

BC = BQ + QC

AB = AP + PB

RC = DR = AP = PB

BQ = QC = QC = BQ  segments on either sides of midpoint

∠BCA = ∠DAB (opposite interior angles of a parallelogram)

ΔRQC ≅ ΔSPA

ΔRCQ = ΔAPS (Side Angle Side SAS rule of congruency)

Therefore, segment QR = segment PS (QR = PS) (Corresponding Parts of Congruent Triangles are Congruent CPCTC)

Similarly we have;

∠CDA = ∠CBA (opposite interior angles of a parallelogram)

ΔSRD ≅ ΔQPB (opposite interior angles of a parallelogram)

Therefore, segment PQ = segment RS (PQ = RS) (Corresponding Parts of Congruent Triangles are Congruent CPCTC).

step-by-step explanation:

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y < -3

step-by-step explanation:

-3y - 2 > 7

-3y -2 + 2 > 7+2

-3y > 9

-3y/-3 < 9/-3   (change the inequality sign when multiplying or dividing by a negative number)

y < -3