Intercept Theorem: Problems with Solutions

Solution:
The lines [tex]AB[/tex] and [tex]CD[/tex] are parallel, so by the intercept theorem, we have [tex]\frac{OD}{OB}=\frac{CD}{AB}[/tex], or [tex]CD=\frac{OD}{OB} \times AB=\frac{12}{3} \times 5=20[/tex].

Solution:
The lines [tex]AB[/tex] and [tex]CD[/tex] are parallel, so by the intercept theorem, we have [tex]\frac{OC}{OA}=\frac{CD}{AB}[/tex], or [tex]CD=\frac{OC}{OA} \times AB=\frac{8}{5} \times 5=8[/tex].

Solution:
The lines [tex]AB[/tex] and [tex]CD[/tex] are parallel, so by the intercept theorem, we have [tex]\frac{OA}{OB}=\frac{OC}{OD}[/tex], or [tex]OC=\frac{OA}{OB} \times OD=\frac{2}{5} \times 15=6[/tex].

Solution:
The lines [tex]AB[/tex] and [tex]CD[/tex] are parallel, so by the intercept theorem, we have [tex]\frac{OB}{BD}=\frac{OA}{AC}[/tex], or [tex]OB=\frac{OA}{AC} \times BD=\frac{5}{3} \times 6=10[/tex].

Solution:
The lines [tex]AB[/tex] and [tex]CD[/tex] are parallel, so by the intercept theorem, we have [tex]\frac{OB}{BD}=\frac{OA}{AC}[/tex], or [tex]OB=\frac{OA}{AC} \times BD=\frac{6}{3} \times 1=2[/tex]. [tex]OD=OB+BD=2+1=3[/tex].

Solution:
The lines [tex]AB[/tex] and [tex]CD[/tex] are parallel, so by the intercept theorem, we have [tex]\frac{OD}{OB}=\frac{CD}{AB}[/tex], or [tex]OD=\frac{CD}{AB} \times OB=\frac{8}{6} \times 12=16[/tex]. [tex]BD=OD-OB=16-12=4[/tex].

Solution:
The parallel lines [tex]AB[/tex], [tex]MN[/tex] and [tex]PQ[/tex] cut the sides of [tex]\angle ACB[/tex], so by the intercept theorem they cut proportional segments on them:
[tex]CQ:QN:NB=CP:PM:MA=3:4:5[/tex]
[tex]\frac{QN}{BC}=\frac{4}{3+4+5}=\frac{1}{3}\Rightarrow BC=3\times 6=18[/tex]

Solution:
Since [tex]ABCD[/tex] is a parallelogram, [tex]BQ\parallel CD[/tex] and [tex]CD=AB=56[/tex].
The lines [tex]QD[/tex] and [tex]BC[/tex] meet at [tex]P[/tex] and are cut by the parallel lines [tex]BQ[/tex] and [tex]CD[/tex], so by the intercept theorem
[tex]\frac{BQ}{CD}=\frac{BP}{PC}=\frac{5}{7}\Rightarrow BQ=\frac{5}{7}\times 56=40[/tex]

Solution:
Let [tex]AB=c[/tex], [tex]BC=a[/tex] and [tex]CA=b[/tex].
The segments [tex]MN[/tex], [tex]NP[/tex] and [tex]PM[/tex] are midsegments, so [tex]MN=\frac{b}{2}[/tex], [tex]NP=\frac{c}{2}[/tex], [tex]PM=\frac{a}{2}[/tex]. Also [tex]AM=MB=\frac{c}{2}[/tex], [tex]BN=NC=\frac{a}{2}[/tex], [tex]CP=PA=\frac{b}{2}[/tex].
So each of the four triangles has sides [tex]\frac{a}{2}[/tex], [tex]\frac{b}{2}[/tex], [tex]\frac{c}{2}[/tex] and its perimeter is half the perimeter [tex]P[/tex] of [tex]\triangle ABC[/tex]:
[tex]4\times \frac{P}{2}=24\Rightarrow P=12[/tex]

Solution:
The point of a median that divides it in the ratio [tex]2:1[/tex] from the vertex is the centroid, so [tex]M[/tex] is the centroid of [tex]\triangle ABC[/tex].
All three medians pass through [tex]M[/tex], so [tex]AK[/tex] and [tex]BP[/tex] are medians: [tex]K[/tex] and [tex]P[/tex] are the midpoints of [tex]BC[/tex] and [tex]AC[/tex]. Then [tex]KP[/tex] is a midsegment of [tex]\triangle ABC[/tex]:
[tex]KP=\frac{AB}{2}=\frac{15}{2}=7.5[/tex]

Solution:
A point equidistant from the sides of an angle lies on its bisector, so [tex]AM[/tex] is the bisector of [tex]\angle BAC[/tex]. By the angle bisector theorem
[tex]\frac{CM}{MB}=\frac{AC}{AB}=\frac{2}{4}=\frac{1}{2}[/tex]
Let [tex]CM=x[/tex]. Then [tex]MB=3-x[/tex] and
[tex]\frac{x}{3-x}=\frac{1}{2}\Rightarrow 2x=3-x\Rightarrow x=1[/tex]

Solution:
Since [tex]H[/tex] lies on [tex]AB[/tex], [tex]BO[/tex] is the bisector of [tex]\angle HBC[/tex] in [tex]\triangle HBC[/tex]. By the angle bisector theorem
[tex]\frac{CO}{OH}=\frac{BC}{BH}\Rightarrow \frac{5}{3}=\frac{BC}{12}\Rightarrow BC=20[/tex]

Solution:
Let [tex]CK[/tex] be the median to [tex]AB[/tex]. The centroid divides it in the ratio [tex]CM:MK=2:1[/tex], so [tex]\frac{CM}{CK}=\frac{2}{3}[/tex].
In [tex]\angle ACK[/tex] we have [tex]PM\parallel AK[/tex], so by the intercept theorem [tex]\frac{CP}{CA}=\frac{CM}{CK}=\frac{2}{3}[/tex]. Since [tex]PQ\parallel AB[/tex],
[tex]\frac{PQ}{AB}=\frac{CP}{CA}=\frac{2}{3}\Rightarrow PQ=\frac{2}{3}\times 9=6[/tex]

Solution:
Let [tex]BL=x[/tex]; then [tex]CL=a-x[/tex]. By the angle bisector theorem
[tex]\frac{BL}{CL}=\frac{AB}{AC}\Rightarrow \frac{x}{a-x}=\frac{c}{b}[/tex]
[tex]bx=c(a-x)\Rightarrow x(b+c)=ac\Rightarrow BL=\frac{ac}{b+c}[/tex]

Solution:
Draw through [tex]M[/tex] the line parallel to [tex]AB[/tex]; it meets the leg [tex]BC[/tex] at a point [tex]N'[/tex]. The legs are cut by the parallel lines [tex]AB[/tex], [tex]MN'[/tex] and [tex]CD[/tex], so by the intercept theorem
[tex]\frac{BN'}{N'C}=\frac{AM}{MD}=\frac{7}{2}[/tex]
On the other hand
[tex]\frac{BN}{NC}=\frac{5.6}{1.6}=\frac{7}{2}[/tex]
Both points lie on the segment [tex]BC[/tex] and divide it in the same ratio [tex]7:2[/tex] counted from [tex]B[/tex], so [tex]N'=N[/tex] and [tex]MN\parallel AB[/tex].