Connect AF. Because DF is the median vertical line of AD and △FAD is an isosceles triangle, FA=FD ∠ADF=∠DAF (or prove that these two triangles are congruent).

So the square of FD = the square of FB =FB*FC.

So FA/FB=FC/FA, so just prove △FAB∽△FCA.

∠∠DAF =∠DAC+∠CAF∠BAD+∠ABF =∠ADF∠ADF =∠DAF∠BAD =∠DAC

∴∠CAF=∠ABF

≈AFC =≈BFA

∴△FAB∽△FCA

∴FA/FB=FC/FA

∴ the square of FA =FB*FC

∴ the square of FD =FB*FC

You should understand, right? You can ask again if you don't understand.