In RT △ ABC, ∠ ACB = 90 °, ab = 5, AC = 3, D is a point on the hypotenuse AB, through point a make AE vertical CD, perpendicular foot e, AE intersection line BC at point F 1 when Tan angle BCD = half, find the length of line BF 2 when f is on the edge BC, let ad be x and BF be y, and find the functional relation of Y with respect to X 3 when BF = 5 / 4, find the length of segment ad More detailed proof!

In RT △ ABC, ∠ ACB = 90 °, ab = 5, AC = 3, D is a point on the hypotenuse AB, through point a make AE vertical CD, perpendicular foot e, AE intersection line BC at point F 1 when Tan angle BCD = half, find the length of line BF 2 when f is on the edge BC, let ad be x and BF be y, and find the functional relation of Y with respect to X 3 when BF = 5 / 4, find the length of segment ad More detailed proof!

1. Because the triangle FCE is similar to FCA, the angle BCD is equal to the angle fac, so the tan angle BCD is equal to the tan angle fac, that is, FC divided by AC equals 1 / 2, so FC = 1.5, so BF = 4-1.5 = 2.52

It is known that the height of the hypotenuse of the RT triangle ABC is CD, e is a point on BC, and the circle AE made by passing through C, e and D intersects F. it is proved that the angle DFE = the angle BAC

In a circle, both DFE and DCE are the circumferential angles of the chord De,
So DFE = angle DCE, that is DFE = angle DCB,
And triangle ACB is similar to triangle CDB, so angle DCB = angle BAC,
So angle DFE = angle BAC

As shown in the figure, in △ ABC, the bisector ad of ∠ ACB = 90 ° intersects BC at point D, CE ⊥ AB intersects ad at point F, AB at point E, DH ⊥ AB at point H
The edge shape cdhf is rhombic

∵∠ ACB = 90 °, bisector ad, DH ⊥ ab of ∠ bac
∴CD=DH,∠CDA=∠HDA
∵CE⊥AB,DH⊥AB
∴CE∥DH
∴∠CFD=∠HDA
∴CF=CD ,CF∥DH
The CFHD is a parallelogram
∵CF=CD
The CFHD is rhombic