In the adjoining figure, ABCD is a trapezium in which AB || DC and P, Q are the midpoints of AD and BC respectively.
Question:
In the adjoining figure, ABCD is a trapezium in which AB || DC and P, Q are the midpoints of AD and BC respectively. DQ and AB when produced meet at E. Also, AC and PQ intersect at R. Prove that (i) DQ = QE, (ii) PR || AB, (iii) AR = RC.
Solution:
Given: AB || DC, AP = PD and BQ = CQ
(i) In ∆QCD and ∆QBE, we have:
∠DQC = ∠BQE (Vertically opposite angles)
∠DCQ = ∠EBQ (Alternate angles, as AE || DC)
BQ = CQ (P is the midpoints)
∴ ∆QCD ≅ ∆QBE
Hence, DQ = QE (CPCT)
(ii) Now, in ∆ADE, P and Q are the midpoints of AD and DE, respectively.
∴ PQ || AE
⇒ PQ || AB || DC
⇒ AB || PR || DC
(iii) PQ, AB and DC are the three lines cut by transversal AD at P such that AP = PD.
These lines PQ, AB, DC are also cut by transversal BC at Q such that BQ = QC.
Similarly, lines PQ, AB and DC are also cut by AC at R.
∴ AR = RC (By intercept theorem)
BQ = CQ (P is the midpoints)
∴ ∆QCD ≅ ∆QBE
Hence, DQ = QE (CPCT)
(ii) Now, in ∆ADE, P and Q are the midpoints of AD and DE, respectively.
∴ PQ || AE
⇒ PQ || AB || DC
⇒ AB || PR || DC
(iii) PQ, AB and DC are the three lines cut by transversal AD at P such that AP = PD.
These lines PQ, AB, DC are also cut by transversal BC at Q such that BQ = QC.
Similarly, lines PQ, AB and DC are also cut by AC at R.
∴ AR = RC (By intercept theorem)
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