Question

The diagonals of quadrilateral ABCD intersect at O. Prove that `"A(∆ACB)"/"A(∆ACD)" = "BO"/"DO"`.

Answer 1

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

To Prove : `[A(∆"ACB")]/[A(∆"ACD")] = "BO"/"DO"`

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

Consider h as the distance between the two parallel sides,

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

So,

`[A(∆"ACB")]/[A(∆"ACD")] = (1/2 xx "AB" xx "h" )/(1/2 xx "CD" xx "h")`

`=("AB")/("CD")`   ....(2)

Now consider the triangles AOB and COD, in which,

`∠ "AOB" = ∠ "COD"`

`∠ "ABO" = ∠ "ODC"` (alternative angle)

`∠ "BAO" = ∠ "DCA"` (alternative angle)

Therefore, `Δ "ODC" ∼ Δ "OBA"`,

`⇒("AO")/("OC") = ("BO")/("DO")=("AB")/("CD")`

`⇒ ("BO")/("DO") = ("AB")/("CD")`

From equations (1) and (2), we get:

`[A(∆"ACB")]/[A(∆"ACD")] = "BO"/"DO"`

Hence prove that `[A(∆"ACB")]/[A(∆"ACD")] = "BO"/"DO"`.