P is a point in the quadrilateral ABCD, connecting PA, Pb, PC, PD and AC, proving s triangle APC = triangle apb-s triangle APD Parallelogram ABCD

It is proved that extending AP to BC to o, crossing B to AP perpendicular to m, crossing d to AP perpendicular to N, crossing C to BM perpendicular to Q. AD / / BC ∠ Dan = ∠ bomcq / / OM ∠ BOM = ∠ BCQ ∠ and = ∠ CQB = 90ad = CB triangle adn congruent triangle cbqdn = BQ, crossing C to CF to fqcfm is rectangle CF = qncf = QN = bm-bq = bm-dns

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