Proving that one of the angles is twice the other I have the following question with me:
"Let ABC be an isosceles triangle with $AB=AC$. Let $D$ be a point on $BC$ such that $BD = 2DC$ and let $P$ be a point on $AD$ such that angle $BAC$ = angle $BPD$. Prove that angle $BAC$ = $2*$angle$(CPD)$"
I know I can use sine and cosine rules and some lengthy calculations and prove it but can anyone help me with an elegant solution for this?
 A: 
Extend $PD$ to point $J$ so that $PB=PJ$. Triangle $PBJ$ is isosceles, like tirangle $ABC$. Because $\angle BAC=\angle BPJ$, base angles must be all equal: 
$$\angle ABC=\angle ACB=\angle PBJ=\angle PJB$$ 
Angles $\angle ACB$ and $\angle AJB$ over segment $AB$ are equal so quadrilateral $ABJC$ is cyclic. Also note that $AB=AC$ ant therefore the angles $\angle AJB$ and $\angle AJC$ are also equal.
In other words, lines $JC$ and $JB$ are symmetric with respect to line $JA$. 
Now construct point $C'$ symmetric to $C$ with respect to line $AJ$. Because of symmetry: $C'\in JB$, $DC=DC'$ and $\angle C'PJ=\angle CPJ$. 
Find midpoint of segment $BD$ and denote it with $K$. We know that $BD=2DC$ and obviously: $BK=KD=DC=DC'$.
So triangle $KDC'$ must be isosceles with $\angle C'KD=\angle KC'D$. But:
$$\angle C'KD+\angle KC'D=2\angle KC'D=\angle C'DC=2\angle C'DJ$$
or:
$$\angle KC'D=\angle C'DJ$$
This means that lines $JD$ and $C'K$ must be parallel. But $K$ is the misdpoint of BD so $C'$ is the midpoint of $BJ$, which is also the base of isosceles triangle BPJ. This simply means that $PC'$ is the line of symmetry of triangle $BPJ$:
$$\angle BPJ=2\angle C'PJ=2\angle CPJ$$
or:
$$\angle BAC=2\angle CPD$$ 
BTW, excellent problem, kept me at bay for quite some time. It certainly deserved more upvotes.
EDIT: Interesting remark: this also means that $\angle PCJ=90^\circ$ and $PCJC'$ is a symmetric cyclic quadrilateral. 
