Sol. Let $R=a\cap b$, $S=c\cap d$ and denote $v=\angle RAB=\angle RBA$, $w=\angle SCD=\angle SDC$. To prove that $KLMN$ is concyclic, it suffices to show that $$\frac{\mathrm{pow}((P),K)}{\mathrm{pow}((Q),K)}=\frac{\mathrm{pow}((P),L)}{\mathrm{pow}((Q),L)}=\frac{\mathrm{pow}((P),M)}{\mathrm{pow}((Q),M)}=\frac{\mathrm{pow}((P),N)}{\mathrm{pow}((Q),N)}$$which is, by Power of a point, equivalent to $$\frac{AK^2}{CK^2}=\frac{AL^2}{DL^2}=\frac{BM^2}{CM^2}=\frac{BN^2}{DN^2}\Leftrightarrow\frac{AK}{CK}=\frac{AL}{DL}=\frac{BM}{CM}=\frac{BN}{DN}$$Note that by law of sines in $\bigtriangleup AKC$, we get $\frac{AK}{CK}=\frac{\mathrm{sin}w}{\mathrm{sin}v}$. Similarly for the other ratios, so we have established the wanted equality. Thus we get $KLMN$ cyclic and $(P)$, $(Q)$ and $(KLMN)$ are coaxial. Let $PQ$ intersect $(KLMN)$ in $H$ like in the diagram. This gives $$\frac{P^2-r_1^2}{Q^2-r_2^2}=\frac{\mathrm{pow}((P),H)}{\mathrm{pow}((Q),H)}=\frac{\mathrm{sin}^2(w)}{\mathrm{sin}^2(v)}$$Let $X$ be the intersection of $PR$ with $(P)$. Now by Law of sines in $\bigtriangleup AXB$, we have $$\mathrm{sin}(v)=\frac{AB}{2r_1}$$Similarly, $\mathrm{sin}(w)=\frac{CD}{2r_2}$ so $$\frac{PH^2-r_1^2}{QH^2-r_2^2}=\frac{\mathrm{sin}^2(w)}{\mathrm{sin}^2(v)}=\frac{r_1^2}{r_2^2}$$which is equivalent to $\frac{PH}{QH}=\frac{r_1}{r_2}$ where we have used that if $\frac{a}{b}=\frac{c}{d}$ then $\frac{a}{b}=\frac{a+c}{c+d}$. Now this implies that $H$ is the center of homothety of $(P)$ and $(Q)$ so the external tangent lines of $(P)$ and $(Q)$ must intersect at $H$.

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for the first result the equality of $AB$ and $CD$ is not needed indeed : wehave easily (with directed angles) $\angle LAB=\angle ABN=\angle CBM , \angle LDC=DCM$ thus $LAD\sim BMC $ so $\angle CMN =\angle ALD\implies \angle KMN=\angle KLN$ hence $LKM N$ is cyclic . consider $I,J$ midpoints of $AB,CD$ $\sin \angle IPA=\sin \angle BAL;\sin \angle JQC=\sin \angle DCM$ thus $\frac{r'}{r }=\frac{IA}{r}: \frac{JC}{r'}=\frac{sin \angle IPA}{\sin \angle JQC}=\frac{\sin \angle BAL}{\sin \angle LDC} =\frac{LD}{LA} $ therefore the ratio of powers of $L$ with respect the two circles is $\frac{P_2(L)}{P_1(L)}=\frac{r'^2}{r^2}$ but $S$ the exsimilicenter of the two circles which is the intersection of the commun external tangents have the same ratio then $S$ is also on $KLMN$.

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