# Thread: Help with proving an identity.

1. ## Help with proving an identity.

I recently found this problem in my maths book, and I don't know how I should solve it:

Show that $L^2=(ab+cd)(ac+bd)/(bc+ad)$

I looked in the answers section of the book, and there it said that I should begin by using the law of cosines on the triangles ABD and BDC, and then use the fact that $\cos C= -\cos A$, as $A+C=180$ degrees. Why is that? Have I forgotten about some basic geometric rule or something?

And still, even after using the fact above, I couldn't complete the task. So how do I complete this task?

2. Originally Posted by scounged
I recently found this problem in my maths book, and I don't know how I should solve it:

Show that $L^2=(ab+cd)(ac+bd)/(bc+ad)$

I looked in the answers section of the book, and there it said that I should begin by using the law of cosines on the triangles ABD and BDC, and then use the fact that $\cos C= -\cos A$, as $A+C=180$ degrees. Why is that? Have I forgotten about some basic geometric rule or something?

opposite angles of a quadrilateral inscribed in a circle are supplementary.
law of cosines ...

$L^2 = a^2 + d^2 - 2ad \cos{C}
$

$L^2 = b^2 + c^2 - 2bc \cos{A} = b^2 + c^2 + 2bc \cos{C}$

$a^2 + d^2 - 2ad \cos{C} = b^2 + c^2 + 2bc \cos{C}$

$a^2 + d^2 - b^2 - c^2 = 2bc \cos{C} + 2ad \cos{C}
$

$a^2 + d^2 - b^2 - c^2 = 2\cos{C}(bc + ad)$

$\dfrac{a^2 + d^2 - b^2 - c^2}{bc+ad} = 2\cos{C}$

sub the left side of the above equation into the equation ...

$L^2 = b^2 + c^2 + 2bc \cos{C}$

$L^2 = b^2 + c^2 + \dfrac{bc(a^2 + d^2 - b^2 - c^2)}{bc+ad}$

common denominator ...

$L^2 = \dfrac{(b^2+c^2)(bc+ad) + bc(a^2 + d^2 - b^2 - c^2)}{bc+ad}$

$L^2 = \dfrac{b^3c+ab^2d + bc^3 + adc^2 + a^2bc + bcd^2 - b^3c - bc^3}{bc+ad}$

$L^2 = \dfrac{ab^2d + adc^2 + a^2bc + bcd^2}{bc+ad}$

$L^2 = \dfrac{a^2bc + ab^2d + adc^2 + bcd^2}{bc+ad}$

$L^2 = \dfrac{(ab+cd)(ac+bd)}{bc+ad}$

3. Thanks alot for the quick reply!