In the adjoining figure, ABCD is a trapezium in which AB || DC and P, Q are the midpoints of AD and BC respectively.


In the adjoining figure, ABCD is a trapezium in which AB || DC and PQ 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.


Given: AB || DCAP = 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 ∆ADEP and Q are the midpoints of AD and DE, respectively.
∴ PQ || AE
⇒ PQ || AB || DC 
⇒​ AB || PR || DC

(iii) PQAB and DC are the three lines cut by transversal AD at P such that AP = PD.
These lines 
PQABDC are also cut by transversal BC at Q such that BQ = QC.
​ Similarly, lines PQAB and DC are also cut by AC at R.
∴ AR = RC                   (By intercept theorem)

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Nov. 29, 2022, 5:06 p.m.
nice 👍 and thanks 😊

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