## Online Course Discussion Forum

### Need help on winter campus blue geometry practice P19-21

Need help on winter campus blue geometry practice P19-21

Re: Need help on winter campus blue geometry practice P19-21

What have you tried so far? Remember we can give you better hints if you tell us how/where you got stuck.

Here are some hints to get you started:

• 19: Use Law of Sines on $\triangle ACD$, and $\triangle ABD$ to show that $$\dfrac{\sin \angle BAD}{\sin \angle CAD} = \dfrac{3}{4}.$$ Then use it again in $\triangle AGE$ and $\triangle AFG$ to find $\dfrac{EG}{GF}$. At some point you might want to use the identity $\sin(\theta) = \sin(180^\circ - \theta)$.
• 20: Denote the sides of $\triangle ABC$ by $a$, $b$, and $c$, and use the fact that $[ABC] = [ABE] + [BEC]$ (here $[ABC]$ means area of $ABC$), along with the area formula $\dfrac{1}{2}ac\sin(B)$ to show that $BE = \dfrac{2ac}{a+c} \cos\left(\frac{B}{2}\right)$. Use the Angle Bisector Theorem to find expressions for $BD$ and $AE$ in terms of $a$, $b$, and $c$, and use these to find an expression for $\cos\left(\frac{B}{2}\right)$ in terms of $a$, $b$, and $c$. Then use Law of Cosines and Law of Sines.
• 21: Try to find what are $BD$ and $AD$: Let $K$ be a point on $\overline{AB}$ such that $AD = BK$; show that $ECFK$ is a parallelogram and use this to show that $D$ is the midpoint of $\overline{GK}$.

Re: Need help on winter campus blue geometry practice P19-21

Thanks a lot, Professor! I will try to follow your hints and see if I am able to solve the problems myself.

Happy New year!

Re: Need help on winter campus blue geometry practice P19-21

Let us know if you need additional hints :)

Happy New Year!

Re: Need help on winter campus blue geometry practice P19-21

For 21, I have found cos(B/2) = ((a+c)^2 - b(b-a))/(2a)(b+c), however, I am not sure where to go from here.  It seems any other calculation I do ends up in a messy equation.  Can you provide some help?

Re: Need help on winter campus blue geometry practice P19-21

You meant for 20, right? (Be careful with parenthesis, you were missing a couple).

Using Law of Cosines you can also find that $$b^2 = (a + c)^2 - 4ac\cos^2\left(\frac{B}{2}\right)$$ (you will need the half angle formula for cosine). You can then combine what you got so far and this to solve for $\cos\left(\frac{B}{2}\right)$ (you should get a quadratic equation in $\cos\left(\frac{B}{2}\right)$).