D and E are the points on the sides AB and AC respectively of a ΔABC

1.	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
Answer» 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

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

Answer»

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