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Triangles — Maths STD 10 — Question

Maharashtra BoardEnglish MediumSTD 10MathsTriangles4 Marks
Question
In the given figure, $\text{DB}\perp\text{BC},\text{DE}\perp\text{AB}$ and $\text{AC}\perp\text{BC}.$
Prove that $\frac{\text{BE}}{\text{DE}}=\frac{\text{AC}}{\text{BC}}.$
✓

Answer


In the given figure : $\text{DB}\perp\text{AB},\text{AC}\perp\text{BC}$ and DB || AC
$\therefore\angle\text{DBC}=\angle\text{ACB}$
AB is the transversal
$\therefore\angle\text{DBE}=\angle\text{BAC}$ $\big[\text{Alternate }\angle\text{s}\big]$
In $\triangle\text{DBE}$ and $\triangle\text{ABC}$
$\angle\text{DEB}=\angle\text{ACB}=90^\circ$
$\angle\text{DBE}=\angle\text{BAC}$
$\triangle\text{DBE}\sim\triangle\text{ABC}$ [By AA similarity]
$\Rightarrow\frac{\text{BE}}{\text{DE}}=\frac{\text{AC}}{\text{BC}}$
Hence proved.

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