(i) In the figure’s ΔAEC and ΔADB.
We have ∠AEC =∠ADB = 90 (∵ CE ∠AB and BD ∠AC) and ∠EAC =∠DAB
[Each equal to ∠A]
Therefore by AA-criterion of similarity, we have
ΔAEC ~ ΔADB
(ii) We have
ΔAEC ~ ΔADB [As proved above]
CA/BA = EC/DB
CA/AB = CE/DB
Hence proved.