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  • If triangles ABC and DBC are on the same base BC and AD intersects BC at O <img alt="" src="https://cdn.entrance360.com/media/uploads/2020/07/20/q14.png" style="height:187px; width:1

If triangles ABC and DBC are on the same base BC and AD intersects BC at O

then \frac{\operatorname{ar}(\Delta A B C)}{\operatorname{ar}(\Delta D B C)}

Option: 1

\frac{A O}{D O}


Option: 2

\left (\frac{A O}{D O} \right )^2


Option: 3

\frac{A L}{D M}


Option: 4

Both (a) and (c)


Answers (1)

best_answer

\\\text{In } \triangle A L O\text{ and }\triangle D M O,\text{ we have} \\\\ \angle A L O=\angle D M O=90^{\circ} \\ \\\text{and } \angle A O L=\angle D O M \quad\text { (vert. opp. } \angle_S). \\\\\therefore \quad \triangle A L O \sim \triangle D M O\qquad\text{ [by AA-similarity]} \\\\\Rightarrow \frac{A L}{D M}=\frac{A O}{D O}

\\\Rightarrow \frac{A L}{D M}=\frac{A O}{D O} \\ \\\therefore \quad \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{\frac{1}{2} \times B C \times A L}{\frac{1}{2} \times B C \times D M}=\frac{A L}{D M}=\frac{A O}{D O} \\ \\\text {Hence, } \frac{\operatorname{ar}(\triangle A B C)}{\operatorname{ar}(\triangle D B C)}=\frac{A O}{D O}

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