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Home » In the given figure, ∆ABC and ∆DBC have the same base BC. If AD and BC intersect at O, prove that .
In the given figure, ∆ABC and ∆DBC have the same base BC. If AD and BC intersect at O, prove that \frac{{ar(\Delta ABC)}}{{ar(\Delta DBC)}} = \frac{{AO}}{{DO}}.
CBSE | Class 10 | Exercise-Formative Assessment | Maths | RS Aggarwal | Triangles
In the given figure, ∆ABC and ∆DBC have the same base BC. If AD and BC intersect at O, prove that \frac{{ar(\Delta ABC)}}{{ar(\Delta DBC)}} = \frac{{AO}}{{DO}}.

Answer:

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\begin{array}{l} \frac{{ar(\Delta ABC)}}{{ar(\Delta DBC)}} = \frac{{\frac{1}{2} \times AX \times BC}}{{\frac{1}{2} \times DY \times BC}}\\ \frac{{ar(\Delta ABC)}}{{ar(\Delta DBC)}} = \frac{{AX}}{{DY}}\\ In\Delta ABCand\Delta DBC,\\ \angle AXY = \angle DYO = {90^0}\\ \angle AOX = \angle DOY\\ \Delta AXO \sim \Delta DYO\\ \frac{{AX}}{{DY}} = \frac{{AO}}{{DO}}\\ \frac{{ar(\Delta ABC)}}{{ar(\Delta DBC)}} = \frac{{AX}}{{DY}} = \frac{{AO}}{{DO}}\\ \frac{{ar(\Delta ABC)}}{{ar(\Delta DBC)}} = \frac{{AO}}{{DO}} \end{array}

Hence, proved.

 

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