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 A intersect at R. Prove that (i) DQ = QE, (ii) PR II AB and (iii) AR = RC.

Answer 1

ANSWER:
Given: AB || DC, AP = PD and BQ = CQ
(i) In $△QCDand△QBE$ we have:
and $△QBE$ we have:
$\angle DQC=\angle BQE\left(Verticallyoppositeangles\right)$
$\angle DCQ=\angle EBQ(Alternateangles,asAE\parallel DC)$
BQ = CQ
(P is the midpoints)
∴
$△QCD\cong △QBE$
Hence, DQ = QE
(CPCT)
(ii) Now, in $△ADE$ P and Q are the midpoints of AD and DE, respectively.
$\therefore PQ\parallel AE$
$\to PQ\parallel AB\parallel DC$
$\to AB\parallel PR\parallel DC$

(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)