Answer:
In \[\Delta \text{ }ABD\text{ }and\text{ }\Delta \text{ }CAE,\]
\[\angle ADE\text{ }=\angle AED\text{ }\left[ Angles\text{ }inverse\text{ }to\text{ }rise\text{ }to\text{ }sides\text{ }are\text{ }equal. \right]\]
In this way, \[\angle ADB\text{ }=\angle AEC\]\[[As\text{ }\angle ADB\text{ }+\text{ }\angle ADE\text{ }=\text{ }{{180}^{o~}}and\text{ }\angle AEC\text{ }+\text{ }\angle AED\text{ }=\text{ }{{180}^{o}}]\]
Furthermore, \[A{{D}^{2}}\text{ }=\text{ }BD\text{ }x\text{ }EC\][Given]
\[AD/\text{ }BD\text{ }=\text{ }EC/\text{ }AD\]
\[AD/\text{ }BD\text{ }=\text{ }EC/\text{ }AE\]
Along these lines, \[\vartriangle ABD\text{ }\sim\text{ }\vartriangle CAE\]by \[SAS\]rule for similitude.