As shown in the figure, the known point P is a point on the square ABCD, PE ⊥ BC, PF ⊥ CD, and the perpendicular feet are e and F, respectively There is no picture,

As shown in the figure, the known point P is a point on the square ABCD, PE ⊥ BC, PF ⊥ CD, and the perpendicular feet are e and F, respectively There is no picture,

P is a point PE ⊥ BC, PF ⊥ CD on the diagonal BD of the square ABCD, and the perpendicular feet are e and F, respectively. Proof: PA = EF proof: ∵ PE ⊥ BC, PF ⊥ CD, ∵ C = 90 °≌ quadrilateral PECF is a rectangle, connecting PC, then PC = EF ∵ AB = CB, BD = BD, ∵ abd = ≌ CBD = 45 °≌ ABP ≌ CBP ≌ pa

As shown in the figure, in square ABCD, e is the midpoint of AB, connecting CE, passing through point d to make DF, perpendicular to point O, intersecting CB at point F (1) to find OC: OE (2) if the square area is
4. Find the s quadrilateral AECD

Let AB = 2A, e be the midpoint of AB, so AE = EB = a ∠