Let $O_B$ be the center of $(BMN)$. Let $O_D$ be the center of $(DKL)$. Let $\ell$ be the line parallel to $MN$ and passing through $D$. Let the lperpendicular from $B$ to $MN$ meet $MN$ at $P$, $KL$ at $Q$, and $\ell$ at $R$. Let $O_BO_D$ meet $BD$ at $S$. Let $BD$ meet $MN$ at $X$ and $KL$ at $Y$. Let $AB=BC=CD=DA=a$. The altitude $h=a\sin (\beta + \theta)$. Let $\angle BMN = \beta$ and $\angle BNM = \theta$. By simple angle calculations, $\angle BAD = \beta + \theta \Rightarrow \angle BAC = \angle DAC = \frac {\beta + \theta}{2}$. $\angle MBP = 90^\circ - \beta$. $\angle MBD = \frac {180^\circ - \beta - \theta}2$. $\angle DBP = \left | \frac{\beta - \theta}2\right|$ $\angle BO_BM = 2\theta \Rightarrow \angle MBO_B = 90^\circ-\theta \Rightarrow \angle O_BBD = \left | \frac{\beta - \theta}2\right|$ (In fact, isogonal conjugate of an altitude passes through the circumcenter) Since $\angle BXM = \angle BYL = \angle KYD$ and $\angle YDK = \angle XBM$, we have $\angle YKD = \angle BMX = \beta$. So $\angle KLD = \angle BNM = \theta$. This yields $\triangle DLK \sim \triangle BNM$. Let $BM=x$ and $DK=xk$. $BP=x\sin \beta$, $QR = xk \sin \beta$, $PQ=h$. $BR = BD \cdot \sin \angle DBR = BD \cdot \cos\left(\frac{\beta - \theta}2\right) $. Since $BD= 2a\cdot \sin \left(\frac{\beta + \theta}{2}\right)$, $x\sin\beta + xk\sin \beta + a\sin (\beta + \theta) = 2a\cdot \sin \left(\frac{\beta + \theta}{2}\right) \cdot \cos\left(\frac{\beta - \theta}2\right)$ $ \Rightarrow x(1+k) = 2a\sin\left(\frac{\beta + \theta}{2}\right) \cdot \frac{ \cos\left(\frac{\beta - \theta}2\right) - \cos\left(\frac{\beta + \theta}2\right) }{\sin \beta} $ $\Rightarrow \boxed{x(1+k) = 2a\sin\left(\frac{\beta + \theta}{2}\right) \cdot \frac {2\sin (\beta / 2) \sin (\theta /2)}{\sin \beta}}$. Let $(BMN)$ meet $BD$ at $T$ and $(DKL)$ meet $BD$ at $U$. $\angle TNB = \angle MBT + \angle BNM =\frac {180^\circ - \beta - \theta}2 + \theta = 90^\circ - \left(\frac {\theta - \beta}{2}\right) $. $\frac {BT}{\sin \angle BNT} = \frac {BM}{\sin \angle BNM } \Rightarrow BT = \frac {x\cos \left(\frac {\theta - \beta}{2}\right)}{\sin \theta}$ Similarly, $DU = \frac {xk\cos \left(\frac {\theta - \beta}{2}\right)}{\sin \theta}$. So $BT + DU = x(1+k)\cdot\frac {\cos \left(\frac {\theta - \beta}{2}\right)}{\sin \theta}$ $\Rightarrow BT+DU = 2a\sin\left(\frac{\beta + \theta}{2}\right) \cdot \frac {2\sin (\beta / 2) \sin (\theta /2)}{\sin \beta} \cdot \frac {\cos \left(\frac {\theta - \beta}{2}\right)}{\sin \theta} $ $\Rightarrow BT + DU = \frac{2a\sin\left(\frac{\beta + \theta}{2}\right)\cos \left(\frac {\theta - \beta}{2}\right)}{2\cos(\beta / 2)\cos (\theta/2)} = \frac{BD \cdot \cos \left(\frac {\theta - \beta}{2}\right)}{ \cos \left(\frac {\theta + \beta}{2}\right)+\cos \left(\frac {\theta - \beta}{2}\right) }$ $\frac{BT+DU}{BD} = \frac{\cos \left(\frac {\theta - \beta}{2}\right)}{ \cos \left(\frac {\theta + \beta}{2}\right)+\cos \left(\frac {\theta - \beta}{2}\right) }$ Since $\beta + \theta < 180^\circ \Rightarrow \beta/2 + \theta /2 < 90^\circ \Rightarrow \cos \left(\frac {\theta + \beta}{2}\right)>0$, $\boxed{BT+DU < BD}$ So $|BT|+|TU| + |UD| = BD$. We know $TO_B \perp MN$ and $UO_D \perp KL$. So $O_BO_D$ should intersect $[TU]$. We called this point $S$. So $S \in [TU]$. Since $O_BB = O_BT$, $S$ is outside of $(BMN)$. So $O_BS > O_BB$. Similarly, $O_DS > O_DB$. So $(BMN)$ and $(DLK)$ do not intersect. $\blacksquare$ Let's do the other part. Let $(MAL)$ cut $AC$ at $W$. Let $(NCK)$ cut $AC$ at $Z$. Claim: $AW + CZ > AC$. If our claim is true, then it is obvious that $(ALM)$ and $(NCK)$ intersect each other. We have $WM=WL$, $\angle MAL = \beta + \theta$. $\frac {ML}{\sin (\beta + \theta)} = \frac {MW}{\sin \left(\frac {\beta + \theta}2\right)} $ and by ptolemy, $AW\cdot ML = MW(AM+AL) \Rightarrow AW = \frac {AM+AL}{2\cos \left(\frac{\beta + \theta}{2}\right)}$. Similarly, $CZ = \frac {CN+CK}{2\cos \left(\frac{\beta + \theta}{2}\right)} $. (This is the special case of Ptolemy for an isosceles triangle [or angle bisector]) So $AW + CZ = \frac {AM+ML+CN+CK}{2\cos \left(\frac{\beta + \theta}{2}\right)} $ Similarly, for the first part, we had $BD>BT + DU = \frac {BM+BN+DK+DL}{2\cos \left(\frac{180^\circ - (\beta + \theta)}{2}\right)} = \frac {BM+BN+DK+DL}{2\sin \left(\frac{\beta + \theta}{2}\right)}$ $\Rightarrow 2a\cdot \sin \left(\frac{\beta + \theta}{2}\right) > \frac {BM+BN+DK+DL}{2\sin \left(\frac{\beta + \theta}{2}\right)} $ $\Rightarrow 4a\cdot \sin^2 \left(\frac{\beta + \theta}{2}\right) > BM + BN + DK + DL$ $\Rightarrow 4a\cdot \left ( 1 - \cos^2 \left(\frac{\beta + \theta}{2}\right)\right) > BM + BN + DK + DL$ $\Rightarrow 4a - (BM+BN+DK+DL) > 4a\cdot \cos^2 \left(\frac{\beta + \theta}{2}\right)$ $\Rightarrow AM+ML+CN+CK > 2a\cdot \cos \left(\frac{\beta + \theta}{2}\right) \cdot 2\cos \left(\frac{\beta + \theta}{2}\right)$ $\Rightarrow AM+ML+CN+CK > AC \cdot 2\cos \left(\frac{\beta + \theta}{2}\right)$ $\Rightarrow AW + CZ = \frac {AM+ML+CN+CK}{2\cos \left(\frac{\beta + \theta}{2}\right)}>AC $ $\Rightarrow $ $(ALM)$ and $(NCK)$ intersect each other. $\blacksquare$

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