# problem!!!!

• Oct 2nd 2012, 02:33 AM
geniusgarvil
problem!!!!
Let ABC be an isosceles triangle with AB = AC. Suppose that the angle bisector
of angle B meets AC at D and that BC = BD + AD. Determine angle A.
• Oct 2nd 2012, 10:49 AM
Soroban
Re: problem!!!!
Hello, geniusgarvil!

This reminds me of a classic problem.
Is there a typo?

Quote:

$\displaystyle ABC$ is an isosceles triangle with $\displaystyle AB = AC.$
The bisector of angle $\displaystyle B$ meets $\displaystyle AC$ at $\displaystyle D$
and $\displaystyle BC \,=\, BD\;{\color{red}=}\;AD.$
Determine angle $\displaystyle A.$

Code:

              A               *             / \             / θ \           /    \           /      * D         /    * 2θ\         / θ *      \       / * θ      2θ \     B * - - - - - - - * C
Let $\displaystyle \theta = \angle ABD = \angle DBC$

Since $\displaystyle \Delta ADB$ is isosceles, $\displaystyle \angle A \,=\,\theta$

$\displaystyle \angle BDC$ is an exterior angle to $\displaystyle \Delta ADB.$
. . Hence: $\displaystyle \angle BDC \,=\,2\theta$

Since $\displaystyle \Delta BDC$ is isosceles, $\displaystyle \angle C \,=\,2\theta$

In $\displaystyle \Delta BDC\!:\;\theta + 2\theta + 2\theta \:=\:180^o \quad\Rightarrow\quad 5\theta \:=\:180^o$

Therefore: .$\displaystyle \angle A \:=\:\theta \:=\:36^o$
• Oct 3rd 2012, 02:53 AM
BobP
Re: problem!!!!
Let the base angles be$\displaystyle 2\theta,$ (later, angles in degrees), in which case the angle $\displaystyle BDA=3\theta.$
Using the sine rule in triangles CBD and ABD, we have, respectively,

$\displaystyle \frac{BD}{\sin(2\theta)}=\frac{CB}{\sin(3\theta)},$ in which case $\displaystyle BD=\frac{CB\sin(2\theta)}{\sin(3\theta)},$

and

$\displaystyle \frac{AD}{\sin\theta}=\frac{BD}{\sin(4\theta)},$ so $\displaystyle AD=\frac{BD\sin\theta}{\sin(4\theta)}=\frac{CB\sin \theta\sin(2\theta)}{\sin(4\theta)\sin(3\theta)}$

Substitute into the given equation $\displaystyle AD+BD=CB$, and cancel $\displaystyle CB$ throughout,

$\displaystyle \frac{\sin\theta\sin(2\theta)}{\sin(4\theta)\sin(3 \theta)}+\frac{\sin(2\theta)}{\sin(3\theta)}=1.$

Multiply throuout by $\displaystyle \sin(4\theta)\sin(3\theta),$

$\displaystyle \sin\theta\sin(2\theta)+\sin(2\theta)\sin(4\theta)-\sin(4\theta)\sin(3\theta)=0.$

Now use the trig identity

$\displaystyle \cos A -\cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

$\displaystyle \{\cos(3\theta)-\cos\theta\}+\{\cos(6\theta)-\cos(2\theta)\}-\{cos(7\theta)-\cos\theta\}=0.$

Rearrange,

$\displaystyle \{\cos(\6\theta)+\cos(3\theta)\}-\{\cos(7\theta)+\cos(2\theta)\}=0,$

and now use the identity

$\displaystyle \cos A + \cos B = 2\cos \left( \frac{A+B}{2}\right)\cos \left( \frac{A-B}{2}\right)$

so

$\displaystyle 2\cos(9\theta/2)\cos(3\theta/2)-2\cos(9\theta/2)\cos(5\theta/2)=0,$

$\displaystyle \cos(9\theta/2)\{\cos(3\theta/2)-\cos(5\theta/2)\}=0,$

and therefore

$\displaystyle \cos(9\theta/2)=0,$ since $\displaystyle \cos(3\theta/2)-\cos(5\theta/2)\neq 0,$

Therefore

$\displaystyle 9\theta/2=90,\quad \theta = 20$ in whch case angle $\displaystyle A=100.$
• Oct 8th 2012, 01:54 AM
geniusgarvil
Re: problem!!!!
thanks it is really awesome