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

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:

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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