Question

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Answer 1

Construction : Draw a perpendicular from vertex A to line BC.

$\frac{A(∆ABD)}{A(∆ADC)} \implies \frac{\frac{1}{2}×AX×BD}{\frac{1}{2}×AX×DC} \\ \huge \implies \frac{BD}{DC} \\ \\ \huge \implies \frac{7}{13} (DC \implies BC-BD)$

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