D and E are the points on the sides AB and AC respectively of a ΔABC such that AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC = 5/2 DE
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D and E are the points on the sides AB and AC respectively of a ΔABC such that AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC = 5/2 DE
D and E are the points on the sides AB and AC respectively of a ΔABC such that AD = 8 cm, DB = 12 cm, AE = 6 cm and CE = 9 cm. Prove that BC = 5/2 DE
We have,
\(\frac{AD}{DB}\) = \(\frac{8}{12}\) = \(\frac{2}{3}\)
And, \(\frac{AD}{EC}\) = \(\frac{6}{9}\) = \(\frac{2}{3}\)
Since, \(\frac{AD}{DB}\) = \(\frac{AD}{EC}\)
Then , by converse of basic proportionality theorem.
DE||BC
In Δ ADE and Δ ABC
∠A = ∠A (common)
∠ADE = ∠B (Corresponding angles)
Then, Δ ADE ~ Δ ABC (By AA similarity)
\(\frac{AD}{AB}\) = \(\frac{DE}{BC}\) (Corresponding parts of similar triangle are proportion)
\(\frac{8}{20}\) = \(\frac{DE}{BC}\)
\(\frac{2}{5}\) = \(\frac{DE}{BC}\)
BC = \(\frac{5}2\) DE