In ` Delta ABC, angle ACB = 90^(@) ," seg "CD bot ` side AB and seg CE is angle bisector of `angle ACB`
Prove : ` (AD)/(BD) = (AE^(2))/(BE^(2))`

Explanation

In ` Delta ACB`,
ray CE bisects ` angle ACB`
` :. ` by theorem of angle bisector of a Delta
` (AC)/(CB)= (AE)/(EB)`
Squaring both the sides , we get ,
`(AC^(2))/(CB^(2)) = (AE^(2))/(EB^(2))`
In ` Delta ACB, angle ACB = 90^(@)`
seg CD `bot` hypotenuse AB
` Delta ACB ~ Delta ADC ~ Delta CDB`
... (Similarity of right angled Delta ) ... (2)
` Delta ACB ~ Delta ADC " " ` ... [ From (2)]
` (AC)/(AD) = (AB)/(AC) " "` (Corresponding sides of similar triangfle )
` :. AC^(2) = AB xx AD " "` ...(3)
also, ` Delta ACB ~ Delta CDB " "` ... [ From (2)]
` :. (AB)/(BC) = (BC)/(BD)" "` ( Correesponding sides of similar Delta )
` :. BC^(2) = AB xx BD " "` ... (4)
Substituting the values of (3) and (4) in (1) , we get
` (ABxx AD)/(ABxxBD) = (AE^(2))/(EB^(2))`
` :. (AD)/(BD) = (AE^(2))/(EB^(2))`