problem!!!!

**geniusgarvil** · October 2nd 2012, 02:33 AM

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.

**Soroban** · October 2nd 2012, 10:49 AM

Hello, geniusgarvil!

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

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

Code:

              A
              *
             / \
            / θ \
           /     \
          /       * D
         /     * 2θ\
        / θ *       \
       / * θ      2θ \
    B * - - - - - - - * C

Let $\theta = \angle ABD = \angle DBC$

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

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

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

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

Therefore: . $\angle A \:=\:\theta \:=\:36^o$

**BobP** · October 3rd 2012, 02:53 AM

Let the base angles be $2\theta,$ (later, angles in degrees), in which case the angle $BDA=3\theta.$
Using the sine rule in triangles CBD and ABD, we have, respectively,

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

and

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

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

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

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

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

Now use the trig identity

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

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

Rearrange,

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

and now use the identity

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

so

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

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

and therefore

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

Therefore

$9\theta/2=90,\quad \theta = 20$ in whch case angle $A=100.$

**geniusgarvil** · October 8th 2012, 01:54 AM

thanks it is really awesome

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