Question
In $\triangle\text{ABC},$ D and E are points on sides AB and AC respectively such that AD × EC = AE × DB. Prove that DE || BC.
✓
Answer
Given: In $\triangle\text{ABC},$ and E are points on sides AB and AC such that AD × EC = AE × DB To Prove: DE || BC 
Proof:
Since AD × EC = AE × DB $\Rightarrow\frac{\text{DB}}{\text{AD}}=\frac{\text{EC}}{\text{AE}}$ $\Rightarrow\frac{\text{DB}}{\text{AD}}+1=\frac{\text{EC}}{\text{AE}}+1$ $\Rightarrow\frac{\text{DB}+\text{AD}}{\text{AD}}=\frac{\text{EC}+\text{AE}}{\text{AE}}$ $\Rightarrow\frac{\text{AB}}{\text{AD}}=\frac{\text{AC}}{\text{AE}}$ $\therefore\text{DE}||\text{BC}$ (Converse of basic proportionality theorem)
Proof:Since AD × EC = AE × DB $\Rightarrow\frac{\text{DB}}{\text{AD}}=\frac{\text{EC}}{\text{AE}}$ $\Rightarrow\frac{\text{DB}}{\text{AD}}+1=\frac{\text{EC}}{\text{AE}}+1$ $\Rightarrow\frac{\text{DB}+\text{AD}}{\text{AD}}=\frac{\text{EC}+\text{AE}}{\text{AE}}$ $\Rightarrow\frac{\text{AB}}{\text{AD}}=\frac{\text{AC}}{\text{AE}}$ $\therefore\text{DE}||\text{BC}$ (Converse of basic proportionality theorem)
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