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.

Answer 1

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)