In `squareABCD` , side `BC||` side `AD`. Digonals `AC` and `BD` intersect each other at `P`. If `AP=(1)/(3)AC` then prove `DP=(1)/(2)BP`.

In squareABCD," side "BC|| side AD. Diagonal AC and diagonal BD intersects in point Q. If AQ=(1)/(3)AC, then show that DQ=(1)/(2)BQ.

Attempt any Two of the following: In squareABCD ,seg AB ||seg CD . Diagonal AC and BD intersect each other at point P . Prove : (A(DeltaABP))/(A(DeltaCPD))=(AB^(2))/(CD^(2))

In square ABCD , seg AD|| seg BC. Diagonal AC and digonal BD intersect each other in point P. Then show that (AP)/(PD)=(PC)/(BP) .

In square ABCD , seg AD|| seg BC. Diagonal AC and digonal BD intersect each other in point P. Then show that (AP)/(PD)=(PC)/(BP) .

In squareABCD, "seg " AD||"seg " BC. Diagonal AC and diagonal BC intersect each other in point P. Then show that (AP)/(PD)=(PC)/(BP)

In ABCD side BC | side AD.Seg AC and seg BD intersect in point Q.If AQ=1/3 AC then show that DQ=(1)/(2)BQ

In a trapezium ABCD, side AB is parallel to side DC, and the diagonals AC and BD intersect each other at point P. Prove that : DeltaAPB is similar to DeltaCPD .

In a trapezium ABCD, side AB is parallel to side DC, and the diagonals AC and BD intersect each other at point P. Prove that : PA xx PD = PB xx PC .

In the figure seg AC and seg BD intersects each other at point P and (AP)/(CP)=(BP)/(DP) . Then Prove that DeltaABP~DeltaCDP .

In the figure, seg AC and seg BD intersect each other in point P and (AP)/(CP)=(BP)/(DP) . Prove that DeltaABP~DeltaCDP .