**Question:**

The diagonals of quadrilateral $\mathrm{ABCD}$ intersect at $\mathrm{O}$. Prove that $\frac{\operatorname{ar}(\Delta \mathrm{ACB})}{\operatorname{ar}(\Delta \mathrm{ACD})}=\frac{\mathrm{BO}}{\mathrm{DO}}$.

**Solution:**

We are given the following quadrilateral with *O* as the intersection point of diagonals

To Prove: $\frac{\operatorname{ar}(\Delta A C B)}{\operatorname{ar}(\triangle A C D)}=\frac{B O}{D O}$

Given *ACB* and *ACD* are two triangles on the same base *AC*

Consider *h* as the distance between two parallel sides

Now we see that the height of these two triangles *ACB* and *ACD* are same and are equal to *h*

So

$\frac{\operatorname{ar}(A C B)}{\operatorname{ar}(A C D)}=\frac{\frac{1}{2} \times A B \times h}{\frac{1}{2} \times C D \times h}$

=\frac{A B}{C D}...(1)

Now consider the triangles *AOB* and *COD* in which

$\angle A O B=\angle C O D$

$\angle A B O=\angle O D C$(alternative angle)

$\angle B A O=\angle D C A$(alternative angle)

Therefore, $\triangle O D C \sim \triangle O B A$

$\Rightarrow \frac{A O}{O C}=\frac{B O}{D O}=\frac{A B}{C D}$

$\Rightarrow \frac{B O}{D O}=\frac{A B}{C D}$...(2)

From equation (1) and (2) we get

$\frac{a r(A C B)}{a r(A C D)}=\frac{B O}{D O}$

Hence prove that $\frac{a r(\triangle A C B)}{a r(\triangle A C D)}=\frac{B O}{D O}$

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