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04/06/2021 11:09 am
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In the figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that area (ΔABC)/area (ΔDBC) = AO/DO.
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04/06/2021 11:21 am
We know that ABC and DBC are two triangles on the same base BC.
AD intersects BC at O.
Area (ΔABC)/Area (ΔDBC) = AO/DO
Let us draw two perpendiculars AP and DM on line BC.
We know that area of a triangle = \(\frac{1}{2}\) × Base × Height
∴ \(\frac{ar(\Delta ABC)}{ar(\Delta ABC)}\)
= \(\frac{\frac{1}{2}BC \times AP}{\frac{1}{2}BC \times DM}\)
= \(\frac{AP}{DM}\)
In ΔAPO and ΔDMO,
∠APO = ∠DMO (Each 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ΔAPO ~ ΔDMO (AA similarity criterion)
∴ AP/DM = AO/DO
⇒ Area (ΔABC)/Area (ΔDBC) = AO/DO.
This post was modified 4 years ago by Raavi Tiwari
This post was modified 3 years ago 2 times by admin