In the right triangle, ∠ ACB = 90 °, CD is high, e is the middle point of BC, the extended line of ED and the extended line of Ca intersect at point F, and AC: BC = DF: CF is calculated

In the right triangle, ∠ ACB = 90 °, CD is high, e is the middle point of BC, the extended line of ED and the extended line of Ca intersect at point F, and AC: BC = DF: CF is calculated

prove:
, ACB = 90 ° and CD is high
∠B=∠ACD
E is the midpoint of BC
∠B=∠BDE=∠ADF
∠FAD=∠FDA
∠F=∠F
Therefore, triangle ADF is similar to triangle DCF
DA/CD=DF/CF
tan∠ACD=AD/CD=tan∠B=AC/BC
So:
AC:BC=DF:CF