Intercept Theorem: Difficult Problems with Solutions

Solution:
For the internal point [tex]AM+MB=12[/tex] and [tex]AM:MB=5:1[/tex], so [tex]MB=2[/tex] and [tex]AM=10[/tex].
For the external point [tex]AN\gt NB[/tex], so [tex]N[/tex] lies on the extension of [tex]AB[/tex] beyond [tex]B[/tex]. Let [tex]NB=x[/tex]; then [tex]AN=12+x[/tex] and
[tex]\frac{12+x}{x}=5\Rightarrow x=3[/tex]
The points [tex]M[/tex] and [tex]N[/tex] lie on different sides of [tex]B[/tex], hence
[tex]MN=MB+BN=2+3=5[/tex]

Solution:
We have [tex]AC=AE+CE=10[/tex]. Since [tex]LE\parallel AB[/tex], the intercept theorem for [tex]\angle ACB[/tex] gives
[tex]\frac{BL}{LC}=\frac{AE}{EC}=\frac{6}{4}=\frac{3}{2}[/tex]
By the angle bisector theorem
[tex]\frac{AB}{AC}=\frac{BL}{LC}=\frac{3}{2}\Rightarrow AB=\frac{3}{2}\times 10=15[/tex]

Solution:
[tex]BM[/tex] is the bisector of [tex]\angle B[/tex], so by the angle bisector theorem
[tex]\frac{AM}{MC}=\frac{AB}{BC}=\frac{10}{15}=\frac{2}{3}\Rightarrow CM=\frac{3}{5}\times 15=9[/tex]
In the same way, from the bisector [tex]AN[/tex] we get [tex]CN=9[/tex]. Since [tex]\frac{CM}{CA}=\frac{CN}{CB}=\frac{3}{5}[/tex], by the converse of the intercept theorem [tex]MN\parallel AB[/tex]. Then
[tex]\frac{MN}{AB}=\frac{CM}{CA}=\frac{3}{5}\Rightarrow MN=\frac{3}{5}\times 10=6[/tex]

Solution:
Draw the diagonal [tex]AC[/tex]; let it meet [tex]PQ[/tex] at [tex]R[/tex]. In [tex]\triangle ACD[/tex] we have [tex]PR\parallel DC[/tex], so by the intercept theorem [tex]\frac{AR}{AC}=\frac{AP}{AD}=\frac{1}{3}[/tex] and
[tex]PR=\frac{1}{3}CD=3[/tex]
In [tex]\triangle ABC[/tex] we have [tex]RQ\parallel AB[/tex] and [tex]\frac{CR}{CA}=\frac{2}{3}[/tex], so
[tex]RQ=\frac{2}{3}AB=12[/tex]
[tex]PQ=PR+RQ=3+12=15[/tex]

Solution:
Let [tex]AC=BC=12[/tex], let [tex]CH[/tex] be the altitude ([tex]H[/tex] is the midpoint of [tex]AB[/tex]), let [tex]P[/tex] be the midpoint of [tex]CH[/tex], and let the line through [tex]P[/tex] parallel to [tex]AC[/tex] meet [tex]AB[/tex] at [tex]E[/tex] and [tex]BC[/tex] at [tex]F[/tex].
In [tex]\triangle AHC[/tex] the line [tex]PE[/tex] passes through the midpoint of [tex]CH[/tex] and is parallel to [tex]CA[/tex], so [tex]E[/tex] is the midpoint of [tex]AH[/tex] and
[tex]BE=BH+HE=\frac{1}{2}AB+\frac{1}{4}AB=\frac{3}{4}AB[/tex]
In [tex]\triangle ABC[/tex] we have [tex]EF\parallel AC[/tex], so by the intercept theorem
[tex]\frac{EF}{AC}=\frac{BE}{BA}=\frac{3}{4}\Rightarrow EF=\frac{3}{4}\times 12=9[/tex]

Solution:
Let the perpendicular bisector meet [tex]AB[/tex] at its midpoint [tex]O[/tex]. Both [tex]MO[/tex] and [tex]CH[/tex] are perpendicular to [tex]AB[/tex], so [tex]MO\parallel CH[/tex]. By the intercept theorem for [tex]\angle ABC[/tex]
[tex]\frac{BO}{BH}=\frac{BM}{BC}=\frac{5}{7}[/tex]
Let [tex]BO=5x[/tex]; then [tex]BH=7x[/tex] and [tex]AB=2BO=10x[/tex], so [tex]AH=AB-BH=3x[/tex] and
[tex]AH:HB=3x:7x=3:7[/tex]

Solution:
Let [tex]G[/tex] be the centroid and [tex]BB_1[/tex] the median to [tex]AC[/tex]; then [tex]\frac{BG}{BB_1}=\frac{2}{3}[/tex]. Since [tex]DE\parallel AC[/tex], the intercept theorem gives
[tex]\frac{BD}{BA}=\frac{BE}{BC}=\frac{DE}{AC}=\frac{BG}{BB_1}=\frac{2}{3}[/tex]
Hence [tex]AD=\frac{AB}{3}[/tex] and [tex]EC=\frac{BC}{3}[/tex], so
[tex]\frac{AB+BC}{3}=16\Rightarrow AB+BC=48\Rightarrow AC=75-48=27[/tex]
[tex]DE=\frac{2}{3}\times 27=18[/tex]

Solution:
Let [tex]F[/tex] be the midpoint of [tex]AM[/tex] and draw [tex]MN\parallel BQ[/tex] with [tex]N\in AC[/tex].
In [tex]\triangle BCQ[/tex] the point [tex]M[/tex] is the midpoint of [tex]BC[/tex] and [tex]MN\parallel BQ[/tex], so by the intercept theorem [tex]N[/tex] is the midpoint of [tex]QC[/tex]: [tex]QN=NC[/tex].
In [tex]\triangle AMN[/tex] the point [tex]F[/tex] is the midpoint of [tex]AM[/tex] and [tex]FQ\parallel MN[/tex], so [tex]Q[/tex] is the midpoint of [tex]AN[/tex]: [tex]AQ=QN[/tex]. Hence
[tex]AQ=QN=NC=\frac{AC}{3}[/tex]
[tex]CQ-AQ=\frac{2}{3}AC-\frac{1}{3}AC=\frac{1}{3}AC=6\Rightarrow AC=18[/tex]

Solution:
Let [tex]AN=MB=x[/tex]; then [tex]AM=3-x[/tex]. Since [tex]MN\parallel BC[/tex], by the intercept theorem
[tex]\frac{AM}{MB}=\frac{AN}{NC}\Rightarrow \frac{3-x}{x}=\frac{x}{4}[/tex]
[tex]x^{2}+4x-12=0\Rightarrow (x-2)(x+6)=0[/tex]
The negative root is rejected, so
[tex]AN=2[/tex]

Solution:
From [tex]AF:FC=3:2[/tex] we get [tex]\frac{CF}{CA}=\frac{2}{5}[/tex]. In [tex]\angle ACB[/tex] we have [tex]FL\parallel AE[/tex], so by the intercept theorem
[tex]\frac{CL}{CE}=\frac{CF}{CA}=\frac{2}{5}\Rightarrow CE=\frac{5}{2}\times 16=40\Rightarrow EL=40-16=24[/tex]
In the same angle [tex]EF\parallel AB[/tex], so
[tex]\frac{CE}{CB}=\frac{CF}{CA}=\frac{2}{5}\Rightarrow CB=\frac{5}{2}\times 40=100\Rightarrow BE=100-40=60[/tex]

Solution:
Let [tex]AB=CD=11x[/tex], so [tex]AE=3x[/tex]. The lines [tex]AC[/tex] and [tex]DE[/tex] meet at [tex]K[/tex] and [tex]AE\parallel CD[/tex], so by the intercept theorem
[tex]\frac{AK}{KC}=\frac{AE}{CD}=\frac{3}{11}\Rightarrow AK=\frac{3}{14}\times 28=6[/tex]
Let [tex]BC=AD=5y[/tex], so [tex]CF=2y[/tex]. The lines [tex]AC[/tex] and [tex]DF[/tex] meet at [tex]L[/tex] and [tex]CF\parallel AD[/tex], so
[tex]\frac{CL}{LA}=\frac{CF}{AD}=\frac{2}{5}\Rightarrow LC=\frac{2}{7}\times 28=8[/tex]
[tex]KL=28-6-8=14[/tex]

Solution:
Let [tex]BC=6[/tex], [tex]AB=16[/tex] and [tex]AC=20[/tex]. The smallest angle lies opposite the smallest side, so it is [tex]\angle A[/tex]. Let its bisector meet [tex]BC[/tex] at [tex]L[/tex]:
[tex]\frac{BL}{LC}=\frac{AB}{AC}=\frac{4}{5}\Rightarrow BL=\frac{4}{9}\times 6=\frac{8}{3}[/tex]
The bisector of the exterior angle at [tex]A[/tex] meets the line [tex]BC[/tex] at a point [tex]E[/tex] with [tex]\frac{EB}{EC}=\frac{AB}{AC}=\frac{4}{5}[/tex]. Since [tex]EB\lt EC[/tex], the point [tex]E[/tex] lies on the extension of [tex]CB[/tex] beyond [tex]B[/tex]. With [tex]EB=x[/tex]:
[tex]\frac{x}{x+6}=\frac{4}{5}\Rightarrow x=24[/tex]
[tex]LE=EB+BL=24+\frac{8}{3}=\frac{80}{3}[/tex]

Solution:
Since [tex]BM\parallel DC[/tex], the intercept theorem for the lines [tex]AC[/tex] and [tex]BD[/tex], which meet at [tex]O[/tex], gives
[tex]\frac{OM}{OC}=\frac{OB}{OD}\Rightarrow OM=\frac{OB\times OC}{OD}[/tex]
Since [tex]CP\parallel AB[/tex], similarly
[tex]\frac{OP}{OB}=\frac{OC}{OA}\Rightarrow OP=\frac{OB\times OC}{OA}[/tex]
Therefore
[tex]\frac{OM}{OA}=\frac{OB\times OC}{OA\times OD}=\frac{OP}{OD}[/tex]
The points [tex]M[/tex] and [tex]P[/tex] lie on the sides [tex]OA[/tex] and [tex]OD[/tex] of [tex]\angle AOD[/tex] and cut off proportional segments, so by the converse of the intercept theorem [tex]MP\parallel AD[/tex].

Solution:
In [tex]\triangle ACD[/tex] we have [tex]KL\parallel DC[/tex], so by the intercept theorem
[tex]\frac{KL}{DC}=\frac{AL}{AC}[/tex]
In [tex]\triangle BCD[/tex] we have [tex]PQ\parallel DC[/tex], so
[tex]\frac{PQ}{DC}=\frac{BQ}{BC}[/tex]
In [tex]\triangle ABC[/tex] we have [tex]LQ\parallel AB[/tex], so
[tex]\frac{AL}{AC}=\frac{BQ}{BC}[/tex]
Hence [tex]\frac{KL}{DC}=\frac{PQ}{DC}[/tex], i.e. [tex]KL=PQ[/tex].

Solution:
On the ray [tex]OA[/tex] take the point [tex]C'[/tex] with [tex]OC'=OC[/tex], and on the ray [tex]OB[/tex] the point [tex]D'[/tex] with [tex]OD'=OD[/tex].
Since [tex]\angle C'OD'=\angle COD[/tex] (vertical angles), the triangles [tex]OC'D'[/tex] and [tex]OCD[/tex] are congruent (SAS), so [tex]\angle OC'D'=\angle OCD[/tex]. These are alternate angles, hence [tex]C'D'\parallel CD[/tex]. From the condition
[tex]\frac{OA}{OC'}=\frac{OA}{OC}=\frac{OB}{OD}=\frac{OB}{OD'}[/tex]
so by the converse of the intercept theorem for [tex]\angle AOB[/tex] we get [tex]AB\parallel C'D'[/tex], and therefore [tex]AB\parallel CD[/tex].
If also [tex]AD\parallel BC[/tex], then [tex]ABCD[/tex] would be a parallelogram, its diagonals would bisect each other and [tex]OA:OC=1[/tex], a contradiction. So [tex]ABCD[/tex] is a trapezoid.