Question

ABCD is a parallelogram. P, Q are the mid-points of the side AB and CD respectively, show that DP and BQ trisects AC and are trisected by AC.

Answer 1

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$<b>Introduction:-</b>$ABCD is a parallelogram whose P and Q are the midpoints of sides AB and CD.

$<b>To prove:-</b>$
DP and BQ trisect AC and AS=SR=RC

$<b>Proof:-</b>$ ABCD is a parallelogram,

therefore, AB||CD [opposite sides of parallelogram are parallel]

=>BP||QD [parts of parallel lines are parallel]

& AB= CD[opposite sides of parallelogram are equal]

1/2 AB = 1/2 CD

Therefore, BP=QD [since P and Q are the midpoint of AB and CD]

In BPDQ,

BP||QD & BP=QD

Therefore,BPDQ is a parallelogram

BQ||PD[opposite sides of parallelogram]

RQ||SD & BR||PS[Parts of parallel lines are equal]

In ∆CSP,

Q is the midpoint of DC and RQ||SD

(LINE DRAWN THROUGH MID-POINT OF ONE SIDE OF A TRIANGLE, PARALLEL TO ANOTHER SIDE, BISECTS THE THIRD SIDE)

therefore,R is the midpoint of CS

=>RS=CR-----(i)

In ∆BAR,

P is the midpoint of AB and BR||PS

(LINE DRAWN THROUGH MID-POINT OF ONE SIDE OF A TRIANGLE, PARALLEL TO ANOTHER SIDE, BISECTS THE THIRD SIDE)

therefore,S is the midpoint of AR

=>RS=AS-----(ii)

from (i) & (ii)

RS=CR=AS

THEREFORE,DP AND BQ TRISECT AC


PROVED



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