We have,
Given that:-
ΔABC right angled at point B.and perpendicular from B intersecting AC at point D.
To prove :-
ΔADB∼ΔABC
ΔBDC∼ΔABC
ΔADB∼ΔBDC
Proof:-
InΔBDCandΔABC
∠C=∠C(Common)
∠BDC=∠ABC(Each90o)
Then,
ΔBDC∼ΔABC(ByAAsimilarity)
InΔADBandΔABC
∠A=∠A(Common)
∠ADB=∠ABC(Each90o)
Then,
ΔADB∼ΔABC(ByAAsimilarity)
Hence proved.