Ok! Construct plane $ABPA'B'$, then intersection of this plane and big sphere is a circle, so we conclude $AP\cdot PA'=BP\cdot PB'$ (*). Moreover, intersection with small spheres is a two tangent circles. On this figure it is easy to show that $AB||A'B'$. It follows that $AP:BP=A'P:B'P$ (**). From (*) and (**) we conclude $AA'=BB'$

Or, you could invert wrt $P$ with power $PA\cdot PA'$. The two tangent spheres turn into two parallel planes, and it's clear from here.

grobber wrote: Or, you could invert wrt $P$ with power $PA\cdot PA'$. The two tangent spheres turn into two parallel planes, and it's clear from here. Hmm... In my opinion you use the same arguments...

I see that now, but I hadn't read your message when I posted mine.

A detailed solution: Consider the inversion with pole $P$ fixing the sphere. It sends $X$ to $X'$ and vice-versa for all $X\in\{A,B,C\}$. Since the spheres $(PABC),(PA'B'C')$ are tangent, the planes $(A'B'C'),(ABC)$, which are the images of those spheres respectively, must be parallel, and we're done: on the one hand, we have $PA\cdot PA'=PB\cdot PB'$, and on the other hand we have $\frac{PA}{PA'}=\frac{PB}{PB'}$, so $\{PA,PA'\}=\{PB,PB'\}\Rightarrow AA'=PA+PA'=PB+PB'=BB'$, and we do the same to prove $AA'=CC'$.

Invert at $P$ with power $\sqrt{PA \cdot PA'}$ followed by reflection in $P$. The spheres $\mathcal{S}$ through $A, B, C, P$ and $\mathcal{S}'$ through $A', B', C', P$ are mapped to planes parallel to each other and tangent to the original sphere. It follows that $PA \cdot PA'=PB \cdot PB'$ and $$\frac{PA}{PA'}=\frac{PB}{PB'},$$so $AA'=BB'$. Similarly, $AA'=CC'$ and we may conclude.

Solved with the help of 17 people (mostly with their cameras off and no participation) [asy][asy] draw(circle((-2,0),1)); draw(circle((1,0),2)); pair A,B,C,D,P; A=(-2,1); B=(-2.455111059789108,-0.8904346821608); C=(1,-2); D=(1.910222119578216,1.780869364392161); P=(-1,0); dot(A^^B^^C^^D^^P); label("$A$",A,N); label("$B$",B,S); label("$A'$",C,S); label("$B'$",(1.910222119578216,1.921869364392161),NW); label("$P$",(-0.7,0),E); draw(A--B--C--D--B--C--A--B--C--D--A--B--D); [/asy][/asy] Consider the cross-section going through the plane defined by the two intersecting lines $AA'$ and $BB'.$ By Power of a Point, we have $$PA~ \cdot ~PA' = PB~ \cdot ~PB' ~~(\clubsuit)~~ \implies \frac{PA}{PB}=\frac{PB'}{PA'} ~~(\alpha)~~\implies \triangle PAB \sim \triangle PB'A'.$$From here, we present two solutions: First, note that $$\triangle PAB \sim \triangle PB'A' \implies \angle PAB = \angle PA'B' \qquad (\heartsuit).$$Since $A,B,A',B'$ were on the circumference of a sphere and the cross-section of a sphere is a circle, we have that $ABA'B'$ is a cyclic quadrilateral and that $$\angle PAB = \angle PB'A' \qquad (\spadesuit).$$Together $\heartsuit$ and $\spadesuit$ imply that $$\angle PA'B' = \angle PB'A' \implies PB'=PA' \stackrel{\alpha}\implies PA=PB.$$ Or, we can proceed without angles: $$\triangle PAB \sim \triangle PB'A' \implies AP ~ \cdot ~ B'P=BP ~\cdot ~A'P ~~(\diamondsuit).$$Adding $\clubsuit$ and $\diamondsuit$ and rearranging yields $$PA~ \cdot ~PA' ~+~AP ~ \cdot ~ B'P= PB~ \cdot ~PB' ~+~BP ~\cdot ~A'P \implies PA(PA'~+~PB') = BP(PA'~+~PB') \implies PA = BP \stackrel{\alpha}\implies PB'=PA'.$$ Either way, we obtain $PA'=PB'$ and $PA=PB,$ therefore $$PA+PA'=PB+PB' \implies AA'=BB'.$$We can derive $BB'=CC'$ in the exact same manner, concluding our proof. $\mathbb{Q.E.D.}$

By PoP $|PA|\cdot |PA'|=|PB|\cdot |PB'|$ and by homothety $\frac{|PA|}{|PA'|}=\frac{|PB|}{|PB'|}.$ Hence $|PA|=|PB|\implies |AA'|=|BB'|.$ Analogously $|BB'|=|CC'|.$

JAnatolGT_00 wrote: By PoP $|PA|\cdot |PA'|=|PB|\cdot |PB'|$ and by homothety $\frac{|PA|}{|PA'|}=\frac{|PB|}{|PB'|}.$ Hence $|PA|=|PB|\implies |AA'|=|BB'|.$ Analogously $|BB'|=|CC'|.$ Seems to be what I did except you just made me realize I could've multiplied the first two equations you wrote down and square-rooted.

Do we really need to use inversion?! I guess not. Here another simple approach. [asy][asy] draw(circle((-2,0),1)); draw(circle((1,0),2)); pair A,B,C,D,P; A=(-2,1); B=(-2.455111059789108,-0.8904346821608); C=(1,-2); D=(1.910222119578216,1.780869364392161); P=(-1,0); dot(A^^B^^C^^D^^P); label("$A$",A,N); label("$B$",B,S); label("$A'$",C,S); label("$B'$",(1.910222119578216,1.921869364392161),NW); label("$P$",(-0.7,0),E); draw(A--B--C--D--B--C--A--B--C--D--A--B--D); [/asy][/asy] Draw a common tangent passing from point $P$ and suppose that the tangent line cuts segments $BA'$ and $AB'$, respectively, at point $M$ and $N$. Then, $\angle{PB'A'}=\angle{BAP}=\angle{BPM}=\angle{NPB'}=\angle{PA'B'}=\angle{ABP}$ from which we get that triangles $ABP$ and $A'B'P$ are isosceles, and have the same base angles. From here we get that $AA'=AP+A'P=BP+B'P=BB'$.