# intruiging geometry problem

• Nov 6th 2011, 01:42 AM
avitos
intruiging geometry problem
okay so here's the problem that i found on the net...

"in the picture, AD is angle bisector if AB + AD = CD and AD + AC = BC ,
angle ACB = 20 degree,"
find angle ABC

here's the link to the picture (http://img694.imageshack.us/img694/6562/pb3.png)

and there are hints onto solving this problem as well... but i still can't doooo it.

AB : AC = area(ABD) : area(ADC) = BD : DC
Use this to express BC/AB and hence cos(angle ACB) in
terms of k = AC/AB. Now, what is sin(angle ABC) in terms
of k and sin(angle ACB)?"

TIA :)
• Nov 8th 2011, 01:00 PM
Opalg
Re: intriguing geometry problem
Quote:

Originally Posted by avitos
okay so here's the problem that i found on the net...

"in the picture, AD is angle bisector if AB + AD = CD and AD + AC = BC ,
angle ACB = 20 degree,"
find angle ABC

here's the link to the picture (http://img694.imageshack.us/img694/6562/pb3.png)

and there are hints onto solving this problem as well... but i still can't doooo it.

AB : AC = area(ABD) : area(ADC) = BD : DC
Use this to express BC/AB and hence cos(angle ACB) in
terms of k = AC/AB. Now, what is sin(angle ABC) in terms
of k and sin(angle ACB)?"

Following the hint, let $\displaystyle k = \frac{AC}{AB} = \frac{CD}{BD}$, so that $\displaystyle AC=k.AB$ and $\displaystyle CD=k.BD$. Then

$\displaystyle \frac{BC}{AB} = \frac{BD+CD}{AB} = (1+k)\frac{BD}{AB}$.

But also $\displaystyle \frac{BC}{AB} = \frac{AC+AD}{AB} = k + \frac{CD-AB}{AB} = k-1+\frac{CD}{AB} = k-1+k\frac{BD}{AB}.$

Comparing those two equations, you see that $\displaystyle k-1 = \frac{BD}{AB}$ and hence $\displaystyle \frac{BC}{AB} = k^2-1.$

Thus $\displaystyle AC=k.AB$ and $\displaystyle BC = (k^2-1)AB.$ Now the cosine formula in triangle ABC tells you that

$\displaystyle \cos20^\circ = \frac{AC^2+BC^2-AB^2}{2AC.BC}.$ Divide top and bottom by $\displaystyle AB^2$ to get

$\displaystyle \cos20^\circ = \frac{k^2 + (k^2-1)^2-1}{2k(k^2-1)} = \frac k2$ (after a bit of simplifying). Therefore $\displaystyle k=2\cos20^\circ.$

Finally, the sine rule tells you that $\displaystyle \frac{\sin(ABC)}{AC} = \frac{\sin20^\circ}{AB}$, from which

$\displaystyle \sin(\angle ABC) = k\sin20^\circ = 2\sin20^\circ \cos20^\circ = \sin40^\circ,$ and so angle $\displaystyle ABC = 40^\circ.$ (Whew)