Question
In the given figure, $D B \perp B C ; D E \perp A B$ and $A C \perp B C$. Prove that $\frac{B E}{D E}=\frac{A C}{B C}$
✓
Answer
$B D \| A C$
In $\triangle B E D$ and $\triangle A B C$,
$\angle B E D=\angle A C B \quad\left(\right.$ each $\left.90^{\circ}\right)$
$\angle D B A=\angle B A C \quad$ (Alternate interior angles)
$\Rightarrow \triangle B E D \sim \triangle A C B$ (By AAA rule)
So, $\frac{B E}{A C}=\frac{D E}{B C}$ (Ratio of corresponding sides of similar traingles)
Or, $\frac{B E}{D E}=\frac{A C}{B C}$
Hence Proved.
In $\triangle B E D$ and $\triangle A B C$,
$\angle B E D=\angle A C B \quad\left(\right.$ each $\left.90^{\circ}\right)$
$\angle D B A=\angle B A C \quad$ (Alternate interior angles)
$\Rightarrow \triangle B E D \sim \triangle A C B$ (By AAA rule)
So, $\frac{B E}{A C}=\frac{D E}{B C}$ (Ratio of corresponding sides of similar traingles)
Or, $\frac{B E}{D E}=\frac{A C}{B C}$
Hence Proved.
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