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 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.

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 . Digonals AC and BD intersect each other at P . If AP=(1)/(3)AC then prove DP=(1)/(2)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 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 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 .

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

In trapezium ABCD, seg AB ||seg DC. Seg BD bot seg AD, seg AC bot seg BC. If AD = 15, BC = 15 and AB = 25, then find A ( square ABCD)