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Similarity — Mathematics STD 10 — Question

ICSE BoardEnglish MediumSTD 10MathematicsSimilarity3 Marks
Question
In figure ABC and DBC are two triangles on the same base BC. Prove that
$\frac{\text { Area }(\triangle ABC )}{\text { Area }(\triangle DBC )}=\frac{ AO }{ DO }$.

✓

Answer

In ΔAOL and ΔDOM
∠ALO = ∠DMO, ...(90° each)
∠AOL = ∠DOM,
(Vertically opposite 2 sides)
(Vert. opp-angles)
∴ ΔAOL ∼ ΔDOM
$\therefore \frac{ AL }{ DM }=\frac{ AO }{ DO }$ ...(1)
If two Δ's are similar the ratio between their corresponding sides is the same.
Now, $\frac{\operatorname{area}(\triangle ABC )}{\operatorname{area}(\triangle DBC )}=\frac{\frac{1}{2} \times BC \times AL }{\frac{1}{2} \times BC \times Dm }=\frac{ AL }{ DM }$
From (1), we get
$\frac{\operatorname{area}(\triangle ABC )}{\operatorname{area}(\triangle DBC )}=\frac{ AO }{ DO }$
Hence proved.

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