Cross-Sections, Angles and Distances in Space: Very Difficult Problems with Solutions

Solution:
Since [tex]A_1B\parallel D_1C[/tex] and [tex]BD\parallel B_1D_1[/tex], the planes [tex](A_1BD)[/tex] and [tex](B_1CD_1)[/tex] are parallel; [tex]BD[/tex] lies in the first and [tex]B_1C[/tex] in the second. The common perpendicular of the skew lines [tex]BD[/tex] and [tex]B_1C[/tex] is perpendicular to two non-parallel lines parallel to these planes, hence to the planes themselves, so the required distance equals the distance between the two planes.
The projection of the diagonal [tex]AC_1[/tex] onto [tex](ABCD)[/tex] is [tex]AC[/tex] and [tex]AC\perp BD[/tex], so [tex]AC_1\perp BD[/tex] by the theorem of the three perpendiculars; similarly its projection onto [tex](ABB_1A_1)[/tex] is [tex]AB_1[/tex] and [tex]AB_1\perp A_1B[/tex], so [tex]AC_1\perp A_1B[/tex]. Therefore [tex]AC_1\perp(A_1BD)[/tex] and also [tex]AC_1\perp(B_1CD_1)[/tex].
Let [tex]AC_1[/tex] meet these planes at [tex]K[/tex] and [tex]L[/tex]; the required distance is [tex]KL[/tex]. The segment [tex]AK[/tex] is the height of the pyramid [tex]AA_1BD[/tex] to its face [tex]A_1BD[/tex], an equilateral triangle with side [tex]a\sqrt{2}[/tex]. We compute the volume [tex]V[/tex] of this pyramid with base [tex]ABD[/tex] and height [tex]AA_1[/tex]:
[tex]V=\frac{1}{3}\times \frac{a^2}{2}\times a=\frac{a^3}{6},\quad S_{A_1BD}=\frac{(a\sqrt{2})^2\sqrt{3}}{4}=\frac{a^2\sqrt{3}}{2}[/tex]
[tex]AK=\frac{3V}{S_{A_1BD}}=\frac{a}{\sqrt{3}}=\frac{a\sqrt{3}}{3}[/tex]
In the same way (pyramid [tex]C_1B_1CD_1[/tex]) [tex]C_1L=\frac{a\sqrt{3}}{3}[/tex]. Since [tex]AC_1=a\sqrt{3}[/tex],
[tex]KL=AC_1-AK-C_1L=a\sqrt{3}-2\times \frac{a\sqrt{3}}{3}=\frac{a\sqrt{3}}{3}[/tex]

Solution:
Let the line [tex]BM[/tex] meet the line [tex]CD[/tex] at [tex]X[/tex]. The right triangles [tex]MAB[/tex] and [tex]MDX[/tex] are congruent ([tex]AM=MD[/tex] and the angles at [tex]M[/tex] are vertical), so [tex]DX=AB=a[/tex] and [tex]XC=2a[/tex].
The points [tex]X[/tex] and [tex]N[/tex] lie both in [tex]\lambda[/tex] and in the plane [tex](CDD_1C_1)[/tex], so [tex]\lambda[/tex] meets this face along [tex]XN[/tex] and [tex]P=XN\cap DD_1[/tex]. From the similar triangles [tex]XDP[/tex] and [tex]XCN[/tex]:
[tex]\frac{DP}{CN}=\frac{XD}{XC}=\frac{1}{2}\Rightarrow DP=\frac{a}{4}[/tex]
The plane meets the faces [tex]ABCD[/tex], [tex]BCC_1B_1[/tex], [tex]CDD_1C_1[/tex], [tex]ADD_1A_1[/tex] along [tex]BM[/tex], [tex]BN[/tex], [tex]NP[/tex], [tex]PM[/tex], so the section is the quadrilateral [tex]BNPM[/tex]. Its orthogonal projection on [tex](ABCD)[/tex] is the quadrilateral [tex]BCDM[/tex] ([tex]N[/tex] and [tex]P[/tex] project to [tex]C[/tex] and [tex]D[/tex]):
[tex]S_{BCDM}=a^2-S_{ABM}=a^2-\frac{a^2}{4}=\frac{3a^2}{4}[/tex]
Let [tex]\varphi[/tex] be the angle between [tex]\lambda[/tex] and [tex](ABCD)[/tex]. Draw [tex]CT\perp BM[/tex] ([tex]T[/tex] on the line [tex]BM[/tex]). Since [tex]NC\perp (ABCD)[/tex], by the theorem of the three perpendiculars [tex]NT\perp BM[/tex], so [tex]\angle NTC=\varphi[/tex]. From [tex]S_{BCM}=\frac{1}{2}BM\times CT=\frac{a^2}{2}[/tex] and [tex]BM=\frac{a\sqrt{5}}{2}[/tex] we get [tex]CT=\frac{2a}{\sqrt{5}}[/tex].
In the right triangle [tex]NCT[/tex] (right angle at [tex]C[/tex]):
[tex]NT=\sqrt{CT^2+CN^2}=\sqrt{\frac{4a^2}{5}+\frac{a^2}{4}}=\frac{a\sqrt{21}}{2\sqrt{5}},\qquad \cos\varphi=\frac{CT}{NT}=\frac{4}{\sqrt{21}}[/tex]
The area of an orthogonal projection equals the area of the figure multiplied by [tex]\cos\varphi[/tex], hence
[tex]S_{BNPM}=\frac{S_{BCDM}}{\cos\varphi}=\frac{3a^2}{4}\times \frac{\sqrt{21}}{4}=\frac{3a^2\sqrt{21}}{16}[/tex]

Solution:
The lines [tex]DN[/tex] and [tex]MC[/tex] are skew. In the base plane draw through [tex]N[/tex] the line [tex]l\parallel MC[/tex], and let [tex]\beta[/tex] be the plane through [tex]DN[/tex] and [tex]l[/tex]. Then [tex]MC\parallel\beta[/tex], so the required distance equals the distance from the point [tex]C[/tex] of [tex]MC[/tex] to the plane [tex]\beta[/tex].
The median [tex]CM[/tex] of the equilateral triangle is also an altitude, so [tex]BM\perp CM[/tex] and [tex]BM=2\sqrt{2}[/tex]. Let [tex]K\in l[/tex] be the foot of the perpendicular from [tex]C[/tex] to [tex]l[/tex]. Since [tex]l\parallel CM[/tex], the segment [tex]CK[/tex] equals the distance from [tex]N[/tex] to [tex]CM[/tex]; as [tex]N[/tex] is the midpoint of [tex]BC[/tex], this distance is half of [tex]BM[/tex] (a midline of triangle [tex]BCM[/tex]):
[tex]CK=\frac{BM}{2}=\sqrt{2}[/tex]
Since [tex]DC\perp (ABC)[/tex] and [tex]CK\perp l[/tex], by the theorem of the three perpendiculars [tex]DK\perp l[/tex]. Hence [tex]l\perp (DCK)[/tex], so the plane [tex]\beta[/tex], which contains [tex]l[/tex], is perpendicular to the plane [tex](DCK)[/tex] and meets it along [tex]DK[/tex].
In triangle [tex]DCK[/tex] draw [tex]CH\perp DK[/tex] ([tex]H\in DK[/tex]); then [tex]CH\perp\beta[/tex], so [tex]CH[/tex] is the required distance. In the right triangle [tex]DCK[/tex] (right angle at [tex]C[/tex]):
[tex]DK=\sqrt{DC^2+CK^2}=\sqrt{2+2}=2,\qquad CH=\frac{DC\times CK}{DK}=\frac{\sqrt{2}\times \sqrt{2}}{2}=1[/tex]

Solution:
The faces [tex]ABB_1A_1[/tex] and [tex]DCC_1D_1[/tex] lie in parallel planes, so the cutting plane meets them along parallel lines: [tex]AK\parallel LC_1[/tex]; in the same way [tex]KC_1\parallel AL[/tex]. Hence [tex]AKC_1L[/tex] is a parallelogram with diagonal [tex]AC_1[/tex], and if [tex]d[/tex] is the distance from [tex]K[/tex] to the line [tex]AC_1[/tex], then
[tex]S=2S_{AKC_1}=AC_1\times d,\qquad AC_1=\sqrt{6^2+3^2+6^2}=9[/tex]
It remains to minimise [tex]d[/tex]. In the base draw [tex]BF\perp AC[/tex], [tex]F\in AC[/tex]. The plane [tex](ACC_1A_1)[/tex] is perpendicular to the base, so [tex]BF\perp (ACC_1A_1)[/tex]. Since [tex]BB_1\parallel AA_1[/tex], the line [tex]BB_1[/tex] is parallel to this plane, and every point [tex]K[/tex] of it is at distance [tex]BF[/tex] from the plane. Hence [tex]d\ge BF[/tex], with equality exactly when the foot [tex]F_1[/tex] of the perpendicular from [tex]K[/tex] to the plane lies on [tex]AC_1[/tex]; this foot is the point above [tex]F[/tex] at height [tex]BK[/tex] ([tex]KF_1\parallel BF[/tex]).
In the right triangle [tex]ABC[/tex] the segment [tex]AF[/tex] is the projection of the leg [tex]AB[/tex] on the hypotenuse:
[tex]AC=\sqrt{36+9}=3\sqrt{5},\quad AF=\frac{AB^2}{AC}=\frac{36}{3\sqrt{5}},\quad \frac{AF}{AC}=\frac{36}{45}=\frac{4}{5}[/tex]
By similar triangles in the plane [tex](ACC_1A_1)[/tex], the point of [tex]AC_1[/tex] above [tex]F[/tex] is at height [tex]CC_1\times \frac{AF}{AC}=6\times \frac{4}{5}=4.8[/tex]. Therefore the area is smallest for
[tex]BK=4.8[/tex]
The diagonals of the parallelogram bisect each other at the center of the box, so [tex]L[/tex] is symmetric to [tex]K[/tex] with respect to it and [tex]DL=6-4.8=1.2[/tex], i.e. [tex]L[/tex] indeed lies on [tex]DD_1[/tex]. (The minimal area is [tex]9\times BF=9\times \frac{6}{\sqrt{5}}=\frac{54\sqrt{5}}{5}[/tex].)

Solution:
Let [tex]AB=x[/tex], [tex]AD=BC=y[/tex], [tex]AA_1=z[/tex]. Each of the three edges is skew to [tex]AC_1[/tex], and the distance between two skew lines equals the distance from one of them to the plane through the other that is parallel to it.
[tex]BB_1\parallel AA_1[/tex], so [tex]BB_1\parallel (ACC_1A_1)[/tex]. This plane is perpendicular to the base, so the distance equals the altitude from [tex]B[/tex] of the right triangle [tex]ABC[/tex]:
[tex]\frac{xy}{\sqrt{x^2+y^2}}=\frac{6}{\sqrt{13}}[/tex]
[tex]BC\parallel AD[/tex], so [tex]BC\parallel (AB_1C_1D)[/tex]. Since [tex]AD\perp (ABB_1A_1)[/tex], this plane is perpendicular to the face [tex]ABB_1A_1[/tex] along [tex]AB_1[/tex], and the distance equals the altitude from [tex]B[/tex] of the right triangle [tex]ABB_1[/tex]:
[tex]\frac{xz}{\sqrt{x^2+z^2}}=\frac{2}{\sqrt{5}}[/tex]
[tex]A_1B_1\parallel AB[/tex], so [tex]A_1B_1\parallel (ABC_1D_1)[/tex]. Since [tex]AB\perp (ADD_1A_1)[/tex], this plane is perpendicular to the face [tex]ADD_1A_1[/tex] along [tex]AD_1[/tex], and the distance equals the altitude from [tex]A_1[/tex] of the right triangle [tex]AA_1D_1[/tex]:
[tex]\frac{yz}{\sqrt{y^2+z^2}}=\frac{3}{\sqrt{10}}[/tex]
Squaring and inverting the three equalities:
[tex]\frac{1}{x^2}+\frac{1}{y^2}=\frac{13}{36}, \quad \frac{1}{x^2}+\frac{1}{z^2}=\frac{5}{4}, \quad \frac{1}{y^2}+\frac{1}{z^2}=\frac{10}{9}[/tex]
Adding them gives [tex]2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)=\frac{98}{36}[/tex], i.e. [tex]\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{49}{36}[/tex]. Subtracting each equation from this:
[tex]\frac{1}{x^2}=\frac{49}{36}-\frac{10}{9}=\frac{1}{4}, \quad \frac{1}{y^2}=\frac{49}{36}-\frac{5}{4}=\frac{1}{9}, \quad \frac{1}{z^2}=\frac{49}{36}-\frac{13}{36}=1[/tex]
Hence [tex]x=2[/tex], [tex]y=3[/tex], [tex]z=1[/tex], and the volume is [tex]V=xyz=6[/tex].

Solution:
Let the plane [tex]\alpha[/tex] meet [tex]DB[/tex] and [tex]DC[/tex] at [tex]P[/tex] and [tex]Q[/tex]. Since [tex]BC\parallel\alpha[/tex], the plane [tex](DBC)[/tex] meets [tex]\alpha[/tex] along a line parallel to [tex]BC[/tex], so [tex]PQ\parallel BC[/tex], the triangle [tex]DPQ[/tex] is equilateral and the section is the triangle [tex]APQ[/tex]. Put [tex]DP=DQ=PQ=ta[/tex], [tex]0\lt t\le 1[/tex].
Let [tex]K[/tex] be the midpoint of [tex]BC[/tex]; the median [tex]DK[/tex] meets [tex]PQ[/tex] at its midpoint [tex]L[/tex], and [tex]DL=t\times DK[/tex]. The triangles [tex]ADP[/tex] and [tex]ADQ[/tex] are congruent ([tex]AD[/tex] common, [tex]DP=DQ[/tex], [tex]\angle ADP=\angle ADQ=60^{\circ}[/tex]), so [tex]AP=AQ[/tex] and [tex]AL\perp PQ[/tex].
Next, [tex]BC\perp AK[/tex] and [tex]BC\perp DK[/tex], so [tex]BC\perp(ADK)[/tex]; hence [tex]PQ\perp(ADK)[/tex], and the plane [tex]\alpha[/tex], which contains [tex]PQ[/tex], is perpendicular to [tex](ADK)[/tex] and meets it along the line [tex]AL[/tex]. Since [tex]BC\parallel\alpha[/tex], the points [tex]B[/tex] and [tex]K[/tex] are at the same distance from [tex]\alpha[/tex], and it equals the distance [tex]KH[/tex] from [tex]K[/tex] to the line [tex]AL[/tex] (the perpendicular [tex]KH[/tex] to [tex]AL[/tex] in the plane [tex](ADK)[/tex] is perpendicular to [tex]\alpha[/tex]). As the angle between [tex]AB[/tex] and [tex]\alpha[/tex] is [tex]30^{\circ}[/tex],
[tex]KH=AB\sin 30^{\circ}=\frac{a}{2}[/tex]
In the isosceles triangle [tex]ADK[/tex] we have [tex]AK=DK=\frac{a\sqrt{3}}{2}[/tex], [tex]AD=a[/tex], and the height from [tex]K[/tex] falls at the midpoint of [tex]AD[/tex], so
[tex]h_K=\sqrt{\frac{3a^2}{4}-\frac{a^2}{4}}=\frac{a\sqrt{2}}{2}, \quad S_{ADK}=\frac{1}{2}\times a\times \frac{a\sqrt{2}}{2}=\frac{a^2\sqrt{2}}{4}[/tex]
[tex]\cos\angle ADK=\frac{AD/2}{DK}=\frac{a/2}{a\sqrt{3}/2}=\frac{1}{\sqrt{3}}[/tex]
Since [tex]KL=(1-t)DK[/tex], we have [tex]S_{AKL}=(1-t)S_{ADK}=\frac{(1-t)a^2\sqrt{2}}{4}[/tex], and by the law of cosines in the triangle [tex]ADL[/tex]
[tex]AL^2=a^2+\frac{3t^2a^2}{4}-2\times a\times \frac{ta\sqrt{3}}{2}\times \frac{1}{\sqrt{3}}=a^2\left(1-t+\frac{3t^2}{4}\right)[/tex]
From [tex]KH=\frac{2S_{AKL}}{AL}=\frac{a}{2}[/tex] we get
[tex]\frac{(1-t)a\sqrt{2}}{2\sqrt{1-t+\frac{3t^2}{4}}}=\frac{a}{2}\Rightarrow 2(1-t)^2=1-t+\frac{3t^2}{4}\Rightarrow 5t^2-12t+4=0[/tex]
The roots are [tex]t=2[/tex] and [tex]t=\frac{2}{5}[/tex]; only [tex]t=\frac{2}{5}[/tex] satisfies [tex]t\le 1[/tex]. Then
[tex]PQ=\frac{2a}{5}, \quad AL^2=a^2\left(1-\frac{2}{5}+\frac{3}{25}\right)=\frac{18a^2}{25}, \quad AL=\frac{3\sqrt{2}a}{5}[/tex]
[tex]S_{APQ}=\frac{1}{2}\times PQ\times AL=\frac{1}{2}\times \frac{2a}{5}\times \frac{3\sqrt{2}a}{5}=\frac{3\sqrt{2}a^2}{25}[/tex]

Solution:
Since [tex]D_1E_1\parallel DE\parallel AB[/tex], the plane through [tex]AB[/tex] and [tex]D_1E_1[/tex] is well defined. Let [tex]K[/tex], [tex]L[/tex] and [tex]K_1[/tex] be the midpoints of [tex]AB[/tex], [tex]DE[/tex] and [tex]D_1E_1[/tex]. The segment [tex]KL[/tex] passes through the center of the base, [tex]KL\perp AB[/tex], and it equals twice the apothem of the hexagon:
[tex]KL=2\times \frac{3\sqrt{3}}{2}=3\sqrt{3}[/tex]
The point [tex]K_1[/tex] projects onto [tex]L[/tex], so by the theorem of the three perpendiculars [tex]KK_1\perp AB[/tex], and [tex]\varphi=\angle K_1KL[/tex] is the angle between the cutting plane and the base. From the right triangle [tex]KLK_1[/tex] with [tex]LK_1=13[/tex]:
[tex]KK_1=\sqrt{27+169}=14,\qquad \cos\varphi=\frac{KL}{KK_1}=\frac{3\sqrt{3}}{14}[/tex]
The diagonal [tex]CF[/tex] passes through the center parallel to [tex]AB[/tex], i.e. halfway between [tex]AB[/tex] and [tex]DE[/tex], so the plane cuts the lateral edges [tex]CC_1[/tex] and [tex]FF_1[/tex] at their midpoints [tex]P[/tex] and [tex]Q[/tex]. Thus the section is the hexagon [tex]ABPD_1E_1Q[/tex], and its orthogonal projection on the base plane is the whole base [tex]ABCDEF[/tex]:
[tex]S_{ABCDEF}=6\times \frac{3^2\sqrt{3}}{4}=\frac{27\sqrt{3}}{2}[/tex]
By the theorem for the area of an orthogonal projection:
[tex]S=\frac{S_{ABCDEF}}{\cos\varphi}=\frac{27\sqrt{3}}{2}\times \frac{14}{3\sqrt{3}}=63[/tex]

Solution:
Let [tex]d(X)[/tex] be the distance from a point [tex]X[/tex] to [tex]\lambda[/tex]. The point [tex]M\in\lambda[/tex] is the midpoint of [tex]AD[/tex], so [tex]A[/tex] and [tex]D[/tex] lie on different sides of [tex]\lambda[/tex] and [tex]d(D)=d(A)=1[/tex].
The point [tex]K\in\lambda[/tex] lies on [tex]DC[/tex]; the perpendiculars from [tex]D[/tex] and [tex]C[/tex] to [tex]\lambda[/tex] form similar right triangles with the segments [tex]KD[/tex] and [tex]KC[/tex], so [tex]d(C):d(D)=KC:KD=3:2[/tex], i.e. [tex]d(C)=\frac{3}{2}[/tex], and [tex]C[/tex] is on the other side of [tex]\lambda[/tex] from [tex]D[/tex]. The point [tex]N\in\lambda[/tex] is the midpoint of [tex]BC[/tex], so [tex]d(B)=d(C)=\frac{3}{2}[/tex] and [tex]B[/tex] is on the other side from [tex]C[/tex].
Hence [tex]A[/tex] and [tex]C[/tex] lie on one side of [tex]\lambda[/tex], [tex]B[/tex] and [tex]D[/tex] on the other, and [tex]\lambda[/tex] meets the edge [tex]AB[/tex] at a point [tex]T[/tex] with [tex]AT:TB=d(A):d(B)=2:3[/tex]. The section is the quadrilateral [tex]MKNT[/tex]; let [tex]S[/tex] be its area.
The part of the tetrahedron containing [tex]A[/tex] and [tex]C[/tex] splits into the pyramid [tex]AMKNT[/tex] and the tetrahedron [tex]ACKN[/tex]. The latter has the same apex [tex]A[/tex] as [tex]ABCD[/tex] and its base lies in the plane [tex](BCD)[/tex], so its volume is [tex]5\times \frac{S_{CKN}}{S_{CDB}}[/tex]:
[tex]V_{AMKNT}=\frac{1}{3}S\times d(A)=\frac{S}{3},\qquad V_{ACKN}=5\times \frac{CK}{CD}\times \frac{CN}{CB}=5\times \frac{3}{5}\times \frac{1}{2}=\frac{3}{2}[/tex]
In the same way the part containing [tex]B[/tex] and [tex]D[/tex] splits into the pyramid [tex]DMKNT[/tex] and the tetrahedron [tex]DBTN[/tex] (apex [tex]D[/tex], base in the plane [tex](ABC)[/tex]):
[tex]V_{DMKNT}=\frac{1}{3}S\times d(D)=\frac{S}{3},\qquad V_{DBTN}=5\times \frac{BT}{BA}\times \frac{BN}{BC}=5\times \frac{3}{5}\times \frac{1}{2}=\frac{3}{2}[/tex]
The two parts make up the whole tetrahedron:
[tex]\frac{S}{3}+\frac{3}{2}+\frac{S}{3}+\frac{3}{2}=5\Rightarrow \frac{2S}{3}=2\Rightarrow S=3[/tex]

Solution:
Let [tex]H[/tex] be the orthogonal projection of [tex]D[/tex] onto [tex](ABC)[/tex]; then [tex]CH[/tex] and [tex]AH[/tex] are the projections of [tex]CD[/tex] and [tex]AD[/tex]. The line [tex]AB[/tex] of this plane is perpendicular to the oblique [tex]CD[/tex], so by the theorem of the three perpendiculars it is perpendicular to its projection: [tex]CH\perp AB[/tex]. In the same way [tex]BC\perp AD[/tex] gives [tex]AH\perp BC[/tex]. Hence [tex]H[/tex] lies on the altitudes of the triangle [tex]ABC[/tex] through [tex]C[/tex] and through [tex]A[/tex], i.e. [tex]H[/tex] is its orthocenter.
If [tex]H[/tex] is a vertex, say [tex]H=C[/tex], then [tex]AC[/tex] is the projection of [tex]AD[/tex], so [tex]AC\perp BC[/tex] and the orthocenter is again [tex]C=H[/tex].

Solution:
Let [tex]H[/tex] be the orthogonal projection of [tex]D[/tex] onto [tex](ABC)[/tex]. First we show that [tex]H[/tex] is the orthocenter of the triangle [tex]ABC[/tex]. The segments [tex]CH[/tex] and [tex]AH[/tex] are the projections of [tex]CD[/tex] and [tex]AD[/tex]. The line [tex]AB[/tex] of this plane is perpendicular to the oblique [tex]CD[/tex], so by the theorem of the three perpendiculars it is perpendicular to its projection: [tex]CH\perp AB[/tex]. In the same way [tex]BC\perp AD[/tex] gives [tex]AH\perp BC[/tex]. Hence [tex]H[/tex] lies on the altitudes of the triangle [tex]ABC[/tex] through [tex]C[/tex] and through [tex]A[/tex], i.e. [tex]H[/tex] is its orthocenter. (If [tex]H[/tex] is a vertex, say [tex]H=C[/tex], then [tex]AC[/tex] is the projection of [tex]AD[/tex], so [tex]AC\perp BC[/tex] and the orthocenter is again [tex]C=H[/tex].)
Since [tex]H[/tex] is the orthocenter, [tex]BH\perp AC[/tex]. But [tex]BH[/tex] is the projection of [tex]BD[/tex] onto [tex](ABC)[/tex] and [tex]AC[/tex] lies in this plane, so by the theorem of the three perpendiculars [tex]AC\perp BD[/tex], i.e. [tex]CA\perp BD[/tex]. (If [tex]H=B[/tex], then [tex]BD\perp(ABC)[/tex] and [tex]AC\perp BD[/tex] directly.)

Solution:
Since [tex]AB\perp CD[/tex], there is a plane through [tex]CD[/tex] perpendicular to [tex]AB[/tex]; let it meet the line [tex]AB[/tex] at [tex]P[/tex]. Then [tex]CP\perp AB[/tex] and [tex]DP\perp AB[/tex], and the Pythagorean theorem in the right triangles [tex]APC[/tex], [tex]BPC[/tex], [tex]APD[/tex], [tex]BPD[/tex] gives
[tex]AC^2-BC^2=AP^2-BP^2=AD^2-BD^2\Rightarrow AC^2+BD^2=AD^2+BC^2[/tex]
In the same way, the plane through [tex]AD[/tex] perpendicular to [tex]BC[/tex] (it exists because [tex]BC\perp AD[/tex]) meets the line [tex]BC[/tex] at a point [tex]Q[/tex], and
[tex]AB^2-AC^2=BQ^2-CQ^2=BD^2-CD^2\Rightarrow AB^2+CD^2=AC^2+BD^2[/tex]
Combining the two equalities:
[tex]AB^2+CD^2=BC^2+AD^2=AC^2+BD^2[/tex]

Solution:
Replacing [tex]\tau[/tex] by a parallel plane does not change any of the three angles. If [tex](ABC)\parallel\tau[/tex], then [tex]\alpha=\beta=\varphi=0^{\circ}[/tex] and the equality is obvious.
Case [tex]1[/tex]: neither [tex]AC[/tex] nor [tex]BC[/tex] is parallel to [tex]\tau[/tex]. Take a plane [tex]\tau'\parallel\tau[/tex] not passing through [tex]C[/tex]; the lines [tex]CA[/tex] and [tex]CB[/tex] meet it at points [tex]A'[/tex] and [tex]B'[/tex]. The triangle [tex]A'B'C[/tex] lies in the plane [tex](ABC)[/tex] and is right-angled at [tex]C[/tex]. Let [tex]C_0[/tex] be the orthogonal projection of [tex]C[/tex] onto [tex]\tau'[/tex] and [tex]d=CC_0\gt 0[/tex]. Then [tex]C_0A'[/tex] and [tex]C_0B'[/tex] are the projections of [tex]CA'[/tex] and [tex]CB'[/tex], so [tex]\alpha=\angle CA'C_0[/tex], [tex]\beta=\angle CB'C_0[/tex] and, from the right triangles,
[tex]\sin \alpha=\frac{d}{CA'},\quad \sin \beta=\frac{d}{CB'}[/tex]
Let [tex]CH[/tex] be the altitude of the triangle [tex]A'B'C[/tex] to [tex]A'B'[/tex]. If [tex]H\ne C_0[/tex], by the theorem of the three perpendiculars [tex]C_0H\perp A'B'[/tex], so [tex]\angle CHC_0[/tex] is the linear angle of the dihedral angle between [tex](ABC)[/tex] and [tex]\tau'[/tex] and [tex]\sin \varphi=\frac{d}{CH}[/tex]. If [tex]H=C_0[/tex], then [tex](ABC)\perp\tau'[/tex], [tex]\varphi=90^{\circ}[/tex] and again [tex]\sin \varphi=1=\frac{d}{CH}[/tex].
In the right triangle [tex]A'B'C[/tex] we have [tex]CA'\times CB'=A'B'\times CH[/tex] and [tex]A'B'^2=CA'^2+CB'^2[/tex], hence
[tex]\frac{1}{CH^2}=\frac{A'B'^2}{CA'^2\times CB'^2}=\frac{1}{CA'^2}+\frac{1}{CB'^2}[/tex]
Multiplying by [tex]d^2[/tex] gives [tex]\sin ^2\varphi=\sin ^2\alpha+\sin ^2\beta[/tex].
Case [tex]2[/tex]: one of the legs, say [tex]AC[/tex], is parallel to [tex]\tau[/tex]. Then [tex]\alpha=0^{\circ}[/tex], and [tex]BC[/tex] is not parallel to [tex]\tau[/tex] (otherwise [tex](ABC)\parallel\tau[/tex]). Again take a plane [tex]\tau'\parallel\tau[/tex] not passing through [tex]C[/tex]; the plane [tex](ABC)[/tex] meets [tex]\tau'[/tex] in a line [tex]l\parallel AC[/tex], and since [tex]BC\perp AC[/tex], also [tex]BC\perp l[/tex].
Let [tex]B'[/tex] be the common point of [tex]BC[/tex] and [tex]\tau'[/tex] (it lies on [tex]l[/tex]) and [tex]C_0[/tex] the projection of [tex]C[/tex] onto [tex]\tau'[/tex]. By the theorem of the three perpendiculars [tex]C_0B'\perp l[/tex], so [tex]\angle CB'C_0[/tex] is both the angle [tex]\beta[/tex] and the linear angle of the dihedral angle [tex]\varphi[/tex] (if [tex]C_0=B'[/tex], both angles equal [tex]90^{\circ}[/tex]). Therefore
[tex]\sin ^2\varphi=\sin ^2\beta=\sin ^2\alpha+\sin ^2\beta[/tex]

Solution:
Let [tex]O=AC\cap BD[/tex]; [tex]O[/tex] is the midpoint of both [tex]AC[/tex] and [tex]BD[/tex], and [tex]SO[/tex] is the common line of the planes [tex](SAC)[/tex] and [tex](SBD)[/tex]. The segment [tex]A_1C_1[/tex] joins points of the sides [tex]SA[/tex] and [tex]SC[/tex] of the triangle [tex]SAC[/tex], so it meets the segment [tex]SO[/tex] at a point [tex]O_1[/tex] of the cutting plane. In the same way [tex]B_1D_1[/tex] meets [tex]SO[/tex] at a point of the cutting plane. Since [tex]S[/tex] is not in the cutting plane, the line [tex]SO[/tex] has only one common point with it, so [tex]B_1D_1[/tex] also passes through [tex]O_1[/tex].
First we prove that in the triangle [tex]SAC[/tex] we have [tex]\frac{SA}{SA_1}+\frac{SC}{SC_1}=\frac{2\,SO}{SO_1}[/tex]. Through [tex]A[/tex] and [tex]C[/tex] draw lines parallel to [tex]A_1C_1[/tex] and let them meet the line [tex]SO[/tex] at [tex]A_2[/tex] and [tex]C_2[/tex]. By the theorem on proportional segments in the angles [tex]ASO[/tex] and [tex]CSO[/tex]:
[tex]\frac{SA}{SA_1}=\frac{SA_2}{SO_1}, \quad \frac{SC}{SC_1}=\frac{SC_2}{SO_1}[/tex]
The triangles [tex]OAA_2[/tex] and [tex]OCC_2[/tex] are congruent ([tex]OA=OC[/tex], the angles at [tex]O[/tex] are vertical, and [tex]\angle OAA_2=\angle OCC_2[/tex] as alternate angles, since [tex]AA_2\parallel CC_2[/tex]), so [tex]OA_2=OC_2[/tex]. The points [tex]A_2[/tex] and [tex]C_2[/tex] lie on the line [tex]SO[/tex] on opposite sides of [tex]O[/tex] (if [tex]A_1C_1\parallel AC[/tex], then [tex]A_2=C_2=O[/tex]), hence [tex]SA_2+SC_2=2\,SO[/tex]. Adding the two proportions, we get
[tex]\frac{SA}{SA_1}+\frac{SC}{SC_1}=\frac{SA_2+SC_2}{SO_1}=\frac{2\,SO}{SO_1}[/tex]
In the same way, in the triangle [tex]SBD[/tex] ([tex]O[/tex] is the midpoint of [tex]BD[/tex] and [tex]B_1D_1[/tex] passes through [tex]O_1[/tex]) we get
[tex]\frac{SB}{SB_1}+\frac{SD}{SD_1}=\frac{2\,SO}{SO_1}[/tex]
Hence the two sums are equal.