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Question

∆ABC is a right triangle right-angled at A and AD ⊥ BC. Then, BDDC=
(a) ABAC2
(b) ABAC
(c) ABAD2
(d) ABAD

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Solution

Given: In ΔABC, and.

To find: BD: DC



CAD+BAD=90° .....1BAD+ABD=90° .....2 ADB=90°From (1) and (2),CAD=ABD

In ΔADB and ΔADC,

ADB=ADC 90° eachABD=CAD ProvedADB~ADC AA SimilarityCDAD=ACAB=ADBD Corresponding sides are proportional


Disclaimer: The question is not correct. The given ratio cannot be evaluated using the given conditions in the question.

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