In the parallelogram ABCD, ab = KBC, point P is on the diagonal BD, and angle ∠ ape = ∠ bad, PE and BC are compared with point E, and the quantitative relationship between PA and PE is explored

In the parallelogram ABCD, ab = KBC, point P is on the diagonal BD, and angle ∠ ape = ∠ bad, PE and BC are compared with point E, and the quantitative relationship between PA and PE is explored

From triangle ADP to triangle PBE, AP / PE = ad / BP
AB / ad = BP / AB is obtained from the similarity of triangle DAB and triangle APB
According to the above two formulas, AP / PE = ad square / AB square = 1 / (k Square)

In the parallelogram ABCD, the diagonal lines AC and BD intersect at O, BD = 2Ab, and points E and F are the midpoint of OA and BC respectively, connecting be and ef. (1) prove EF = BF

∵ ABCD is a parallelogram, ∵ Bo = BD / 2, BD = 2Ab, ∵ Bo = AB, and BF = Bo / 2,
∴BF＝AB/2.
∵ E and F are the middle points of AO and Bo, respectively, ∵ EF is the median line of △ OAB, ∵ EF = AB / 2, ∵ EF = BF

The diagonal lines AC and BC of parallelogram ABCD intersect at point o.ef, pass through point O, and intersect with AD and BC at points E and f respectively, AE = 6 and BF = 3, so the length of ad can be obtained

Ad = 9, triangle BOF is equal to triangle doe, so BF = De, ad = AE + de = 9

In the parallelogram ABCD, if EF passes through the diagonal intersection o, the proof is de = BF
Like the title,
If AB = 4, ad = 3, of = 1.3, find the perimeter of trapezoid afed

Proof of de = BF by congruent triangle