Question

ABCD is a parallelogram.ACandBD are the diagonals intersect at O. Pand Q are the points of tru section of the diagonal bd. prove that CQ||AP and also AC bisects PQ.

Answer 1

DO=OB.......... (YOU CAN MAKE YOUR FIGURE)
ON SUBTRACTING DP FROM BOTH SIDES
DO - DP=OB-DP
BUT, DP=QB
SO, PO=PQ........ (1) (AC bisects PQ)
AND, AO =OC (diagonal of Parallelogram)
And angle AOP =angle QOC (vertically opposite angles)
So, triangle APO=triangle OQC[SAS]
so, angle APQ =angle QOC
AP parallel to CP

Answer 2

In triangle AOP & QOC.

Angle AOP & QOC are vertically opposite angles .so they are equal.

=> Angle AOP= Angle QOC

we know diagnol bisect each other .so BD divided AC in to two equal parts i.e, AO & OC .

so AO =OC

=> And also DO =BO

=> DP+PO=OQ+QB.........(1)

In the question it is given that,

p & q are trisectional points of BD .so

DP=PQ=BQ

Substitute in equation1

=> DP+PO=OQ+DP

=> PO=DO

By SAS congruence.

These triangles are congruent. corresponding angles are also equal.

PAO=QCO

Interior alternative angles are equal .so ,
AP||QC

In the above it is proved that PO= QO.

Hence AC bisect PO.

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