Answer.
The condition above can be drawn below!
Connect PQRS such that it is drawn a smaller 4-gon.
Look at triangle ABD and PBR!
Obvious BP:BA = BR:BD. So that we can conclude that triangle BPR similar to triangle BAD. Thus, we can conclude that PR // AD ……… (1*)
Look at triangle CAD and CSQ!
Obvious CS:CA = CQ:CD. So that we can conclude that triangle CSQ similar to triangle CAD. Thus, we can conclude that QS // AD ……… (2*)
Because PR // AD and QS // AD, then we conclude that PR // QS ……… (3*)
Look at triangle ABC and APS!
Obvious AP:AB = AS:AC. So that we can conclude that triangle APS similar to triangle ABC. Thus, we can conclude that PS // BC ……… (4*)
Look at triangle DRQ and DBC!
Obvious DR:DB = DQ:DC. So that we can conclude that triangle DRQ similar to triangle DBC. Thus, we can conclude that QR // BC ……… (5*)
Because PS // BC and QR // BC, then we conclude that PS // QR ……… (6*)
Because PR // QS and PS // QR, we can conclude that PQRS is a parallelogram by the definition. And its characteristics, by theorem 1.6, we can see that PQ and RS intersects at midpoint of it.
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