Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ, \angle{BAP}=\angle{BCP} = 20^\circ$. Show $\triangle{ABC}$ is an isosceles Let $P$ be a point inside $\triangle{ABC}$ such that $\angle{ABP}=10^\circ$, $\angle{BAP}=\angle{BCP} = 20^\circ$. Show $\triangle{ABC}$ is an isosceles and find $\angle{CPA}$.
extend $AP$, and tried law of sines; also googled and searched on stackexchange, no clues.
This seems to be one of the hard IMO problems https://davidaltizio.web.illinois.edu/CollectionOfGeometryProblems.pdf or Sharygin geometry problems https://www.isinj.com/mt-aime/Problems%20in%20Plane%20Geometry%20-%20Sharygin%20(MIR,1982).pdf and no good solution yet.
As shown by answers, the original problem doesn't give enough conditions to prove what's asked. The missing part is $\angle PBC=30^\circ$. Then one can make a mirror point $P'$ of $P$ over $BC$, connect $AP'$, $AB=PC$, $\Delta BPC$ is equilateral, and $BAP$ and $P'AP$ are congruent, $\Delta AP'C$ is equilateral, so $AB=PC=AP'=P'C=AC$.

initial conditions: Let $P$ be a point inside $△ABC$ such that $∠ABP=10^\circ$, $∠BAP=∠BCP=20^\circ$
if $C\in  R=λ_{A}∪λ_{B}∪l$ then the triangle ABC is an isosceles.
$\begin{array}{} λ_{A}(A,1) & λ_{B}(B,1) & (l) \text{ perpendicular bisector of the line segment AB.} \end{array}$
If $C$ belongs to the intersection of $R$ with the circle $\lambda'$ then we have the solution satisfying the conditions.
($λ'$ is the reflection of $λ(O,1)$ with respect to line $PB$). $\lambda'$ is locus of C such that $\angle{BCP}={20^\circ}$. Two solutions:$C$ and $C_{1}$.  $O$ is the circumcenter of the triangle $ABP$. $O=(\frac{1}{2},-\frac{\sqrt{3}}{2})$.
straight line $AP$ through $A(0,0)$ with slope equal to $tan (\angle{20}^\circ)$ and $BP$ through $B(1,0)$ with slope equal to $-tan (\angle{10}^\circ)$. ($P=AP∩PB$) ($(l)$ $x=\frac{1}{2}$)
$h=tan\left(  \frac{π}{18}  \right)$
$\begin{array}{} x_{P}=\frac{h^2-1}{h^2-3} & y_{P}=\frac{2h}{3-h^2}  \end{array}$
$\begin{array}{} x_{O´}=\frac{3h^2-5}{2(h^2-3)} & y_{O'}=\frac{\sqrt{3}(h^2-3)-4h}{2(h^2-3)}  \end{array}$
$C=\left( \frac{1}{2},\frac{\sqrt{3}(h^2-3)-4(h+\sqrt{2-h^2})}{2(h^2-3)}  \right)$
$b^2=\frac{[h(8-2\sqrt{3}h)+6\sqrt{3}]\sqrt{2-h^2}+h^3(h-2\sqrt{3})-6h(h-\sqrt{3})+17}{(h^2-3)^2}$
$t^2=\frac{h^2-5-2\sqrt{3}\sqrt{2-h^2}}{h^2-3}$
$d=\frac{h^2+1}{h^2-3}$
$cos(θ)=\frac{-b^2+d^2+t^2}{2dt}$
$θ=cos^{-1}\left( \frac{h^3(h+2\sqrt{3})-2h\left( 3\sqrt{3}+4\sqrt{2-h^2}-1  \right) }{2(h^2+1)\sqrt{15+h^2(h^2-8)-2(h^2-3)\sqrt{6-3h^2}}}  \right)$
$\theta≅115.476252^\circ$
where: $b=AC$, $t=CP$, $d=AP$,  $\theta=\angle{APC}$
$θ_{2}=∠(APC_{1})=80°$

