Since AC is the other diagonal of square ABCD, AC and BD perpendicularly bisect each other at the midpoint of AC, O(4,3). So, OA=OB=OC=OD, and ∠AOB=90°. 1). Thru O, draw a horizontal y=3, and a vertical x=4. From A and C, draw 2 verticals toward the horizontal y=3. Label each foot on the line, E(7,3) and F(1,3), respectively. 2). From B and D, 2 horizontals toward the vertical x=4. Label each foot on the line, G and H, respectively. These ΔAOE, ΔBOG, ΔCOF and ΔDOH are right triangles. 3). In ΔAOE and ΔCOF, AE//CF. So, ∠OAE=OCF (alternate angles), ∠AOE=∠COF (vertical angles) and OA=OC. Thus, ΔAOE≡ΔCOF (A.S.A). In the same way, ΔBOG≡ΔDOH (A.S.A). 4). In ΔAOE and ΔBOG, AC⊥BD and EF⊥GH. So, ∠AOE=90°-∠EOB=∠BOG. OA=OB, OE//BG and ∠OAE=90°-∠AOE=∠EOB=∠OBG (alternate angles). Thus,ΔAOE≡ΔBOG (A.S.A). Therfore, AE=BG=CF=DH=1, OE=OG=OF=OH=3, and the coordinates of B and D are B(5,6) and D(3,0). 5). A linear equation that passes thru 2 points, P(x1,y1) and Q(x2,y2), is expressed as follows: (y-y1) / (y2-y1)=(x-x1) / (x2-x1). To find a equation that passes thru B and D, plug x1=5, y1=6, x2=3 and y2=0 into the equation and simplify it. (y-6) / (0-6)=(x-5) / (3-5), y=3x-9. CK this at point O(4,3). Plug x=4 into the equation. y=3x-9=3x4-9=3. Points B,O and D are on a straight line y=3x-9. CKD. The equation of diagonal BD is y=3x-9 (3≦x≦5).