Q. 105.0( 1 Vote )

In ΔABC, D is the mid-point of and P is the mid-point of. intersects in Q. Prove that (i) CQ = 2AQ (ii) BP = 3PQ

Answer :

Given: In ΔABC,


BD = DC [ D is mid-point of BC]


AP = PD [ P is mid-point of AD]


Construction: Draw a line l from D parallel to AC and let it intersect BQ at S and AB at R.


Draw line m from D parallel to AC intersecting AC at T.


Draw a line from C parallel to BQ.


Extend AD and let it intersect the parallel drawn from C at point E.



Proof:


In ΔADT,


PQ||DT [ by construction]


AP = PD [ Given]


Hence, by mid-point theorem,


AQ = QT …(1)


In ΔBQC,


BD = DC [ given: d is mid-point of BC]


DS|| QC [ by construction]


Hence, by mid-point theorem,


QT = TC …(2)


By eq. (1), (2) and (3) we get that,


AQ = QT =TC …(4)


AQ = 2QC [As QC = QT + TC]


In ΔPDB and ΔCDE,


CDE = BDP [vertically opposite angles]


BD = DC [Given]


PBD = ECD [alternate interior angles between BP||CE]


Therefore, ΔCDE ΔPDB.


CE = BP [By CPCT] …(5)


In ΔAPQ and ΔAEC,


A = A [common angle]


APQ = AEC [Corresponding angles when PQ||CE]


AQP = ACE [Corresponding angles when PQ||CE]


By AAA similarity,


ΔAPQ ΔAEC


And,



[from (4)]


[From (5)]


PQ = 3BP


Hence proved.


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