solve this..........
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Given, ABCD is a parallelogram.
BC = AD and BC||AD
Also, DC = AB and DC||AB
Since, P is mid-point of DC.
DP = PC = 1/2DC
Now, QC||AP and PC||AQ
APCQ is a parallelogram.
 AQ = PC = 1/2DC = 1/2AB = BQ
Now, in ΔAQR and ΔBQC,
BQ = AQ
∠BQC = ∠AQR
And ∠BCQ = ∠ARQ
 ΔAQR BQC
 AR = BC
But BC = DA
 AR = DA
Also, CQ = QR
Hence, Proved
BC = AD and BC||AD
Also, DC = AB and DC||AB
Since, P is mid-point of DC.
DP = PC = 1/2DC
Now, QC||AP and PC||AQ
APCQ is a parallelogram.
 AQ = PC = 1/2DC = 1/2AB = BQ
Now, in ΔAQR and ΔBQC,
BQ = AQ
∠BQC = ∠AQR
And ∠BCQ = ∠ARQ
 ΔAQR BQC
 AR = BC
But BC = DA
 AR = DA
Also, CQ = QR
Hence, Proved
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