Answer:
Given :
ABC is a right triangle in which ∠ABC = 90°,
Also, Legs AB and CB are extended past point B to points D and E,
Such that,

To prove :

Proof :
In triangles AEC and EBA,
∠EAC= ∠ABE ( right angles )
∠CEA = ∠AEB ( common angles )
By AA similarity postulate,
,
Similarly,


Now, In triangles ADC and CBD,
∠ACD = ∠CBD ( right angles )
∠ADC= ∠BDC ( common angles )
By AA similarity postulate,
,
Similarly,


From equations (1) and (2),

The corresponding sides of similar triangles are in same proportion,


Hence, proved....