In ∆ABC, D and E are points on sides AB and AC respectively such that AD ✕ EC = AE ✕ DB. Prove that DE || BC.

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Solution

Given: In, D and E are points on sides AB and AC such that

To Prove:

Proof:

Since

∴ DE || BC (Converse of basic proportionality theorem)

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D and E are points on the sides AB and AC respectively of a $Δ A B C$ such that DE || BC
Find the value of x, when

(i) AD = x cm, DB = (x - 2) cm,

AE = (x + 2) cm and EC = (x - 1) cm.

(ii) AD = 4 cm, DB = (x - 4) cm, AE = 8 cm and EC = (3x - 19) cm.

(iii) AD = (7x - 4) cm, AE = (5x - 2) cm, DB = (3x + 4 ) cm and EC = 3x cm.

If D and E are points on sides AB and AC respectively of $Δ A B C$ such that DE || BC and BD = CE. Prove that $Δ A B C$ is isoscles.

D and E are points on sides AB and AC respectively of ΔABC such that

ar (DBC) = ar (EBC). Prove that DE || BC.

\frac{A D}{A B}

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