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In the given figure, `angle BAC=90^(@) and AD bot BC`.then,

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In the given figure, `angle BAC=90^(@) and AD bot BC`.then,
image
A. `BC*CD=BC^(2)`
B. `AB*AC=BC^(2)`
C. `BD*CD=AD^(2)`
D. `AB*AC=AD^(2)`
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Correct Answer - C
In `Delta DBA and Delta DAC`, we have
`angle ADB= angle CDA=90^(@), angle ABD= angle CAD=90^(@)- angle C`
and `angle BAD=angle ACD=90^(@)- angle B`
`:. Delta DBA~Delta DAC rArr (BD)/(AD)=(AD)/(CD) rArr BD*CD=AD^(2)`
image
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