Intercept Theorem: Very Difficult Problems with Solutions

Solution:
In the right triangle [tex]BDC[/tex] the segment [tex]BF[/tex] bisects [tex]\angle DBC[/tex], so by the angle bisector theorem
[tex]\frac{BC}{BD}=\frac{CF}{FD}=\frac{26}{10}=\frac{13}{5}[/tex]
Let [tex]BD=5k[/tex], [tex]BC=13k[/tex]; also [tex]DC=10+26=36[/tex]. By the Pythagorean theorem
[tex](13k)^{2}-(5k)^{2}=36^{2}\Rightarrow 144k^{2}=1296\Rightarrow k=3[/tex]
So [tex]BD=15[/tex] and [tex]BC=39[/tex]. In the right triangle [tex]ABD[/tex] the segment [tex]BE[/tex] bisects [tex]\angle ABD[/tex], so
[tex]\frac{AB}{BD}=\frac{AE}{ED}=\frac{5}{3}\Rightarrow AB=25, \quad AD=\sqrt{25^{2}-15^{2}}=20[/tex]
[tex]AC=AD+DC=56, \quad P=25+39+56=120[/tex]

Solution:
Let [tex]AB=a[/tex] be the base and [tex]AC=BC=b[/tex] the legs; then [tex]a+2b=32[/tex]. The altitude [tex]CH[/tex] is also a bisector, so the intersection point [tex]I[/tex] of the bisectors lies on [tex]CH[/tex], and [tex]AH=\frac{a}{2}[/tex].
In [tex]\triangle AHC[/tex] the segment [tex]AI[/tex] bisects [tex]\angle A[/tex], so
[tex]\frac{CI}{IH}=\frac{AC}{AH}=\frac{2b}{a}\Rightarrow \frac{CI}{CH}=\frac{2b}{2b+a}=\frac{2b}{32}=\frac{b}{16}[/tex]
Let the line through [tex]I[/tex] meet [tex]AC[/tex] and [tex]BC[/tex] at [tex]M[/tex] and [tex]N[/tex]. By the intercept theorem
[tex]\frac{MN}{AB}=\frac{CM}{CA}=\frac{CI}{CH}=\frac{b}{16}\Rightarrow MN=\frac{ab}{16}=6\Rightarrow ab=96[/tex]
Substituting [tex]a=32-2b[/tex]:
[tex]b^{2}-16b+48=0\Rightarrow b_1=4, \ b_2=12[/tex]
For [tex]b=4[/tex] we get [tex]a=24\gt 2b[/tex], which violates the triangle inequality. Hence [tex]b=12[/tex] and the base is
[tex]a=32-24=8[/tex]

Solution:
The line [tex]AB[/tex] passes through the center and through the point [tex]M[/tex] of the circle, so it meets the circle again at a point [tex]L[/tex] such that [tex]ML[/tex] is a diameter. Then [tex]\angle MCL=90^{\circ}[/tex], i.e. [tex]CL\perp CM[/tex], so [tex]CL[/tex] is the bisector of the exterior angle at [tex]C[/tex]. By the angle bisector theorem
[tex]\frac{AC}{BC}=\frac{AM}{MB}=\frac{12}{3}=4[/tex]
Since [tex]AC\gt BC[/tex], the exterior bisector meets the line [tex]AB[/tex] beyond [tex]B[/tex], and by the exterior angle bisector theorem [tex]\frac{AL}{LB}=\frac{AC}{BC}=4[/tex]. With [tex]LB=x[/tex]:
[tex]\frac{15+x}{x}=4\Rightarrow x=5[/tex]
[tex]ML=MB+BL=3+5=8[/tex]

Solution:
Let [tex]BM[/tex] be the median ([tex]M[/tex] is the midpoint of [tex]AC[/tex]) and [tex]G[/tex] the centroid, so [tex]BG:GM=2:1[/tex]. Let [tex]BL[/tex] ([tex]L\in AC[/tex]) be the bisector of [tex]\angle B[/tex]; the incenter [tex]I[/tex] lies on it, and [tex]L\neq M[/tex] (otherwise [tex]GI[/tex] would lie on [tex]BM[/tex]).
Since [tex]GI\parallel ML[/tex], the intercept theorem for [tex]\angle MBL[/tex] gives
[tex]\frac{BI}{IL}=\frac{BG}{GM}=2[/tex]
Denote [tex]BC=a[/tex], [tex]AC=b[/tex] and [tex]AB=c[/tex]. From the bisector [tex]BL[/tex]: [tex]\frac{AL}{LC}=\frac{c}{a}[/tex], so [tex]AL=\frac{bc}{a+c}[/tex]. In [tex]\triangle ABL[/tex] the segment [tex]AI[/tex] bisects [tex]\angle A[/tex], so
[tex]\frac{BI}{IL}=\frac{AB}{AL}=\frac{c(a+c)}{bc}=\frac{a+c}{b}[/tex]
Hence
[tex]\frac{a+c}{b}=2\Rightarrow a+c=2b=2\Rightarrow a+b+c=3[/tex]

Solution:
Let [tex]O[/tex] be the intersection point of the diagonals; it is the midpoint of [tex]AC[/tex] and [tex]BD[/tex], so [tex]BO[/tex] is a median of [tex]\triangle ABC[/tex] and its centroid [tex]G_1[/tex] lies on [tex]BO[/tex] with [tex]\frac{BG_1}{BO}=\frac{2}{3}[/tex].
Let the line through [tex]G_1[/tex] parallel to [tex]AC[/tex] meet [tex]AB[/tex] and [tex]BC[/tex] at [tex]P[/tex] and [tex]Q[/tex]. By the intercept theorem
[tex]\frac{BP}{BA}=\frac{BQ}{BC}=\frac{BG_1}{BO}=\frac{2}{3}\Rightarrow PQ=\frac{2}{3}AC=18[/tex]
In the same way the line through the centroid of [tex]\triangle ADC[/tex] meets [tex]CD[/tex] and [tex]DA[/tex] at [tex]R[/tex] and [tex]T[/tex] with [tex]\frac{DR}{DC}=\frac{DT}{DA}=\frac{2}{3}[/tex] and [tex]RT=18[/tex].
Now [tex]\frac{CQ}{CB}=\frac{CR}{CD}=\frac{1}{3}[/tex], so by the converse of the intercept theorem [tex]QR\parallel BD[/tex] and [tex]QR=\frac{1}{3}BD=4[/tex]. Likewise [tex]PT=4[/tex]. The perimeter is
[tex]18+4+18+4=44[/tex]

Solution:
Draw [tex]DP\parallel AK[/tex] with [tex]P\in BC[/tex]. By the intercept theorem for [tex]\angle ACB[/tex]
[tex]\frac{CP}{PK}=\frac{CD}{DA}=\frac{5}{4}[/tex]
so we may write [tex]PK=4t[/tex] and [tex]CP=5t[/tex]. By the intercept theorem for [tex]\angle DBC[/tex] (with [tex]OK\parallel DP[/tex])
[tex]\frac{BK}{KP}=\frac{BO}{OD}=\frac{3}{2}\Rightarrow BK=6t[/tex]
[tex]KC=KP+PC=9t\Rightarrow BK:KC=6t:9t=2:3[/tex]

Solution:
By the angle bisector theorem [tex]\frac{BL}{CL}=\frac{AB}{AC}=2[/tex], so [tex]BL=2CL=AL[/tex]. Thus [tex]\triangle ABL[/tex] is isosceles; let [tex]\angle LAB=\angle LBA=\beta[/tex], then [tex]\angle BAC=2\beta[/tex].
Let [tex]K[/tex] be the midpoint of [tex]AB[/tex]; then [tex]AK=\frac{AB}{2}=AC[/tex]. The triangles [tex]ACL[/tex] and [tex]AKL[/tex] are congruent ([tex]AC=AK[/tex], [tex]\angle CAL=\angle KAL[/tex], [tex]AL[/tex] common), so [tex]\angle ACL=\angle AKL[/tex].
In the isosceles [tex]\triangle ABL[/tex] the median [tex]LK[/tex] is also an altitude, so [tex]\angle AKL=90^{\circ}[/tex] and [tex]\angle C=90^{\circ}[/tex]. Then
[tex]2\beta+\beta=90^{\circ}\Rightarrow \beta=30^{\circ}[/tex]
[tex]\angle A=60^{\circ}, \ \angle B=30^{\circ}, \ \angle C=90^{\circ}[/tex]

Solution:
Let [tex]ABC[/tex] have base [tex]AB[/tex], [tex]CA=CB=30[/tex], and let [tex]CH[/tex] be the altitude ([tex]H[/tex] is the midpoint of [tex]AB[/tex]). Let [tex]G_1[/tex], [tex]G_2[/tex] lie on [tex]CH[/tex] with [tex]CG_1=G_1G_2=G_2H[/tex], and let [tex]AG_1[/tex] and [tex]AG_2[/tex] meet [tex]BC[/tex] at [tex]P[/tex] and [tex]Q[/tex].
Draw [tex]HK\parallel AQ[/tex] with [tex]K\in BC[/tex]. In [tex]\triangle ABQ[/tex] the point [tex]H[/tex] is the midpoint of [tex]AB[/tex], so [tex]K[/tex] is the midpoint of [tex]QB[/tex]: [tex]QK=KB=x[/tex]. In [tex]\angle HCB[/tex] we have [tex]G_2Q\parallel HK[/tex], so by the intercept theorem
[tex]\frac{CQ}{QK}=\frac{CG_2}{G_2H}=2\Rightarrow CQ=2x\Rightarrow 4x=30\Rightarrow CQ=QB=15[/tex]
In the same way draw [tex]HL\parallel AP[/tex] with [tex]L\in BC[/tex]; then [tex]PL=LB=y[/tex] and
[tex]\frac{CP}{PL}=\frac{CG_1}{G_1H}=\frac{1}{2}\Rightarrow CP=\frac{y}{2}\Rightarrow \frac{y}{2}+2y=30\Rightarrow y=12[/tex]
[tex]CP=6, \ PQ=15-6=9, \ QB=15[/tex]

Solution:
Let [tex]CA=b[/tex], [tex]CB=a[/tex] and [tex]CD=h[/tex]. Let the line through [tex]E[/tex] meet [tex]CB[/tex] at [tex]K[/tex], and the line through [tex]F[/tex] meet [tex]CA[/tex] at [tex]L[/tex]. By the intercept theorem in [tex]\angle ACB[/tex]
[tex]\frac{CL}{CA}=\frac{CF}{CB}\Rightarrow CL=\frac{hb}{a}, \quad \frac{CK}{CB}=\frac{CE}{CA}\Rightarrow CK=\frac{ha}{b}[/tex]
So the parts cut off on the sides are
[tex]EL=\frac{h|b-a|}{a}, \quad KF=\frac{h|b-a|}{b}\Rightarrow \frac{EL}{KF}=\frac{b}{a}=\frac{12}{9}=\frac{4}{3}[/tex]
Hence [tex]b\gt a[/tex] and [tex]h\left(\frac{4}{3}-1\right)=12[/tex], i.e. [tex]h=36[/tex]. Let the two lines meet [tex]CD[/tex] at [tex]E_1[/tex] and [tex]F_1[/tex]. By the intercept theorem in [tex]\angle ACD[/tex] and [tex]\angle BCD[/tex]
[tex]CE_1=\frac{CE}{CA}\times CD=\frac{h^{2}}{b}, \quad CF_1=\frac{h^{2}}{a}[/tex]
Both lines are perpendicular to [tex]CD[/tex], so their distance is
[tex]E_1F_1=\frac{h^{2}}{a}-\frac{h^{2}}{b}=\frac{1296}{a}\times \frac{1}{4}=\frac{324}{a}=6\Rightarrow a=54, \quad b=\frac{4}{3}\times 54=72[/tex]

Solution:
In [tex]\triangle ACD[/tex] we have [tex]MO\parallel DC[/tex], so by the intercept theorem [tex]\frac{MO}{b}=\frac{AM}{AD}[/tex]. In [tex]\triangle ABD[/tex] we have [tex]MO\parallel AB[/tex], so [tex]\frac{MO}{a}=\frac{DM}{DA}[/tex]. Adding the two equalities:
[tex]\frac{MO}{a}+\frac{MO}{b}=\frac{AM+MD}{AD}=1\Rightarrow MO=\frac{ab}{a+b}[/tex]
In the same way, from [tex]\triangle BCD[/tex] and [tex]\triangle ABC[/tex] we get [tex]ON=\frac{ab}{a+b}[/tex]. Therefore
[tex]MN=MO+ON=\frac{2ab}{a+b}[/tex]

Solution:
Let [tex]P[/tex] be the midpoint of [tex]AB[/tex]. Through [tex]A[/tex] and [tex]B[/tex] draw lines parallel to [tex]MN[/tex]; let them meet the line [tex]CP[/tex] at [tex]A_1[/tex] and [tex]B_1[/tex]. By the intercept theorem for [tex]\angle MCP[/tex] and [tex]\angle NCP[/tex]
[tex]\frac{CA}{CM}=\frac{CA_1}{CP}, \quad \frac{CB}{CN}=\frac{CB_1}{CP}[/tex]
The triangles [tex]APA_1[/tex] and [tex]BPB_1[/tex] are congruent ([tex]AP=BP[/tex], vertical angles at [tex]P[/tex], and [tex]\angle PAA_1=\angle PBB_1[/tex] as alternate angles), so [tex]PA_1=PB_1[/tex] and the points [tex]A_1[/tex], [tex]B_1[/tex] lie on opposite sides of [tex]P[/tex]. Hence one of the segments [tex]CA_1[/tex], [tex]CB_1[/tex] equals [tex]CP+PA_1[/tex] and the other equals [tex]CP-PA_1[/tex], so
[tex]CA_1+CB_1=2CP\Rightarrow \frac{CA}{CM}+\frac{CB}{CN}=\frac{CA_1+CB_1}{CP}=2[/tex]

Solution:
The lines [tex]AN[/tex] (it contains [tex]M[/tex] and [tex]P[/tex]) and [tex]QR[/tex] (it contains [tex]C[/tex]) meet at [tex]P[/tex]. Since [tex]AB\parallel DC[/tex], these two lines are cut by the parallel lines [tex]AQ[/tex] and [tex]NC[/tex], so by the intercept theorem
[tex]\frac{PQ}{PC}=\frac{PA}{PN}[/tex]
Since [tex]AD\parallel BC[/tex], they are also cut by the parallel lines [tex]AR[/tex] and [tex]MC[/tex], so
[tex]\frac{PR}{PC}=\frac{PA}{PM}[/tex]
Because [tex]P[/tex] is the midpoint of [tex]MN[/tex], [tex]PM=PN[/tex], so the right-hand sides are equal. Hence [tex]\frac{PQ}{PC}=\frac{PR}{PC}[/tex], i.e. [tex]PQ=PR[/tex].

Solution:
Let [tex]A_2[/tex] be the midpoint of [tex]BA_1[/tex] and [tex]B_2[/tex] the midpoint of [tex]AB_1[/tex]. The midsegment of [tex]\triangle ABB_1[/tex] gives [tex]MB_2\parallel BB_1[/tex] and [tex]MB_2=\frac{1}{2}BB_1[/tex]; the midsegment of [tex]\triangle BAA_1[/tex] gives [tex]MA_2\parallel AA_1[/tex] and [tex]MA_2=\frac{1}{2}AA_1[/tex]. Since [tex]AA_1=BB_1[/tex], we get [tex]MA_2=MB_2[/tex].
In [tex]\angle MCB_2[/tex] we have [tex]PB_1\parallel MB_2[/tex], and in [tex]\angle MCA_2[/tex] we have [tex]PA_1\parallel MA_2[/tex], so by the intercept theorem
[tex]\frac{PB_1}{MB_2}=\frac{CP}{CM}=\frac{PA_1}{MA_2}\Rightarrow PA_1=PB_1[/tex]
Then [tex]AP=AA_1-PA_1=BB_1-PB_1=BP[/tex], so [tex]\triangle ABP[/tex] is isosceles and its median [tex]PM[/tex] is also an altitude: [tex]CM\perp AB[/tex]. Thus in [tex]\triangle ABC[/tex] the median [tex]CM[/tex] is also an altitude, so [tex]AC=BC[/tex].