ABC is a right triangle at C,a perpendicular CD is drawn at AB prove AC^2/BC^2=AD/BD.?

in triangle ACD

angle ACD + angle CAD = 90

in triangle ACB

angle CAB + ABC = 90

angle CAB = angle CAD same angle

therefore angle ACD = angle ABC --------------------------------(1)

consider the Triangles

ACD and CBD

angle ACD = angle CBD | angle ABC = angle CBD

angle ADC = angle CDB

so the two triangles are similar

so we have

AD/CD = AC/CB = DC/DB -----------------------------------2)

AD/DC = AC/BC

DC/BD = AC/BC | from 2

multiplying both

AD/DC * DC/BD = AC/BC * AC/BC

AD /BD = AC^2/BC^2 prooved

Find the distance between the points d, e, and f and if the distance is equal between all of them then they make an equilateral triangle.