Similar Triangles: Difficult Problems with Solutions

Solution:
The angles [tex]\angle DCA=\angle CAB[/tex] are alternate angles and correspond in the similarity, so [tex]\frac{DC}{CA}=\frac{CA}{AB}=\frac{DA}{CB}=\frac{8}{12}=\frac{2}{3}[/tex].
Hence [tex]CA=\frac{2}{3} \times 27=18[/tex] and [tex]DC=\frac{2}{3} \times 18=12[/tex].

Solution:
From the right triangles [tex]\triangle BHC[/tex] and [tex]\triangle BDA[/tex]: [tex]BH=BC\cos B[/tex] and [tex]BD=BA\cos B[/tex], so [tex]\frac{BH}{BC}=\frac{BD}{BA}=\cos B[/tex].
The triangles have the common [tex]\angle B[/tex] and the sides at it are proportional, so by the SAS criterion [tex]\triangle DBH \sim \triangle ABC[/tex].

Solution:
[tex]CD=16[/tex], and [tex]AD=BD[/tex] because the triangle is isosceles. Both angles in [tex]\angle HAD=\angle BCD[/tex] complement the angle at [tex]B[/tex] to a right angle, so [tex]\triangle AHD \sim \triangle CBD[/tex] and [tex]\frac{HD}{AD}=\frac{BD}{CD}[/tex].
Hence [tex]AD^2=HD \times CD=9 \times 16=144[/tex], so [tex]AD=12;\ AB=24[/tex], and [tex]AC=\sqrt{12^2+16^2}=20[/tex].

Solution:
[tex]AM=11[/tex]. Both angles in [tex]\angle HBM=\angle CAM[/tex] complement the angle at [tex]C[/tex] to a right angle, so for the right triangles [tex]\triangle BHM \sim \triangle ACM[/tex].
Hence [tex]\frac{HM}{BM}=\frac{CM}{AM}[/tex], so [tex]CM=\frac{6 \times 11}{8}=\frac{33}{4}[/tex] and [tex]BC=8+\frac{33}{4}=\frac{65}{4}[/tex].

Solution:
In [tex]\triangle BCH[/tex], [tex]BD[/tex] bisects the angle at [tex]B[/tex], so [tex]\frac{CD}{DH}=\frac{BC}{BH}=\frac{5}{3}[/tex]. Let [tex]BC=5k;\ BH=3k[/tex]; then [tex]CH=\sqrt{BC^2-BH^2}=4k=24[/tex], so [tex]k=6[/tex] and [tex]BC=30;\ BH=18[/tex].
From [tex]\triangle ABC \sim \triangle CBH[/tex]: [tex]AB=\frac{BC^2}{BH}=\frac{900}{18}=50[/tex].

Solution:
Let [tex]K[/tex] be the point of [tex]BC[/tex] with [tex]C_1K \parallel AN[/tex]. In [tex]\triangle ABN[/tex], [tex]C_1[/tex] is the midpoint of [tex]AB[/tex] and [tex]C_1K \parallel AN[/tex], so [tex]BK=KN[/tex]. In [tex]\triangle CC_1K[/tex], [tex]MN \parallel C_1K[/tex], so [tex]\frac{CN}{NK}=\frac{CM}{MC_1}=\frac{1}{3}[/tex].
Let [tex]CN=x;\ NK=3x;\ BK=3x[/tex]; then [tex]BN=6x[/tex] and [tex]BN:NC=6:1[/tex].

Solution:
Since [tex]AB \parallel CD[/tex], [tex]\triangle AOB \sim \triangle COD[/tex] (vertical and alternate angles), so [tex]\frac{AB}{CD}=\frac{AO}{OC}=3[/tex] and [tex]AB=9[/tex].
The midsegment is [tex]\frac{AB+CD}{2}=\frac{9+3}{2}=6[/tex].

Solution:
Let [tex]BC=a[/tex]. By the angle bisector theorem [tex]\frac{BL}{LC}=\frac{AB}{AC}=\frac{20}{32}[/tex], so [tex]BL=\frac{20a}{52}=\frac{5a}{13}[/tex]; also [tex]BM=\frac{a}{2}[/tex].
Since [tex]AB \lt AC[/tex], [tex]L[/tex] lies between [tex]B[/tex] and [tex]M[/tex], so [tex]LM=\frac{a}{2}-\frac{5a}{13}=\frac{3a}{26}[/tex] and [tex]a=26[/tex].

Solution:
The opposite sides of the rhombus are parallel: [tex]DE \parallel BC[/tex], so [tex]\triangle ADE \sim \triangle ABC[/tex] and [tex]\frac{DE}{BC}=\frac{AD}{AB}[/tex], i.e. [tex]\frac{x}{10}=\frac{15-x}{15}[/tex]. Hence [tex]x=6[/tex].

Solution:
By the angle bisector theorem [tex]\frac{AB}{AC}=\frac{BL}{LC}=\frac{4}{6}[/tex], so [tex]AB=8[/tex], and [tex]BC=10[/tex]. Then [tex]\frac{AK}{KC}=\frac{AB}{BC}=\frac{8}{10}[/tex], so [tex]AK=\frac{8}{18} \times 12=\frac{16}{3}[/tex].
In [tex]\triangle ABK[/tex] the segment [tex]AO[/tex] bisects the angle at [tex]A[/tex], so [tex]\frac{BO}{OK}=\frac{AB}{AK}=8:\frac{16}{3}=\frac{3}{2}[/tex].

Solution:
By the angle bisector theorem [tex]\frac{CM}{MB}=\frac{AC}{AB}=\frac{8}{6}[/tex], so [tex]CM=\frac{4}{7} \times 8=\frac{32}{7}[/tex]. By symmetry [tex]CP=CM[/tex], so [tex]MP \parallel AB[/tex] and [tex]\frac{MP}{AB}=\frac{CM}{CB}=\frac{4}{7}[/tex].
Hence [tex]MP=\frac{24}{7}[/tex].

Solution:
By the angle bisector theorem in [tex]\triangle AMB[/tex]: [tex]\frac{AD}{DB}=\frac{AM}{MB}[/tex], and in [tex]\triangle AMC[/tex]: [tex]\frac{AP}{PC}=\frac{AM}{MC}[/tex]. Since [tex]MB=MC[/tex], [tex]\frac{AD}{DB}=\frac{AP}{PC}[/tex], so by the converse of Thales' theorem [tex]DP \parallel BC[/tex].

Solution:
By the angle bisector theorem [tex]\frac{AQ}{QB}=\frac{CA}{CB}=\frac{2}{3}[/tex] and [tex]\frac{BP}{PC}=\frac{AB}{AC}=\frac{5}{4}[/tex]. Take [tex]CA=4k;\ CB=6k;\ AB=5k[/tex].
So [tex]BC:CA:AB=6:4:5[/tex].

Solution:
The triangles share the angle at [tex]C[/tex] and [tex]\angle DBC=\angle CAB[/tex], so [tex]\triangle CBD \sim \triangle CAB[/tex]: [tex]\frac{CB}{CA}=\frac{CD}{CB}=\frac{BD}{AB}[/tex].
Hence [tex]CB^2=CD \times CA=2 \times 18=36[/tex], so [tex]CB=6[/tex], and [tex]AB=BD \times \frac{CA}{CB}=5 \times 3=15[/tex].

Solution:
[tex]\frac{AD}{AC}=\frac{4}{6}=\frac{2}{3}[/tex] and [tex]\frac{AC}{AB}=\frac{6}{9}=\frac{2}{3}[/tex], and the triangles have the common angle at [tex]A[/tex], so by the SAS criterion [tex]\triangle ADC \sim \triangle ACB[/tex].
Hence [tex]\frac{CD}{BC}=\frac{2}{3}[/tex] and [tex]CD=\frac{2}{3} \times 12=8[/tex].

Solution:
Both angles in [tex]\angle DAM=\angle ABD[/tex] complement [tex]\angle ADB[/tex] to a right angle, and both triangles are right triangles, so [tex]\triangle ADM \sim \triangle BAD[/tex].
Hence [tex]\frac{DM}{AD}=\frac{AD}{AB}[/tex] and [tex]DM=\frac{36}{9}=4[/tex].

Solution:
Since [tex]AB \parallel CD[/tex], [tex]\frac{AO}{OC}=\frac{AB}{CD}=2[/tex]. From [tex]\triangle AOM \sim \triangle ACD[/tex]: [tex]\frac{OM}{CD}=\frac{AO}{AC}[/tex], so [tex]OM=\frac{2}{3} \times 5=\frac{10}{3}[/tex]; in the same way [tex]ON=\frac{10}{3}[/tex], so [tex]MN=\frac{20}{3}[/tex].

Solution:
The external bisector divides the opposite side externally in the ratio of the adjacent sides: [tex]\frac{PB}{PC}=\frac{AB}{AC}=\frac{10}{4}[/tex]. So [tex]PB=\frac{5}{2} \times 6=15[/tex] and [tex]BC=PB-PC=15-6=9[/tex].

Solution:
[tex]AC=6;\ BC=8[/tex], and [tex]AH=\frac{AC^2}{AB}=\frac{36}{10}=\frac{18}{5}[/tex]. Let [tex]K[/tex] be the projection of [tex]G[/tex] on [tex]AB[/tex]. The centroid is the average of the vertices, so its projection is the average of the projections of the vertices, which lie at distances [tex]0;\ AB;\ AH[/tex] from [tex]A[/tex]: [tex]AK=\frac{0+10+\frac{18}{5}}{3}=\frac{68}{15}[/tex].
Since [tex]CH[/tex] is perpendicular to [tex]AB[/tex], the distance from [tex]G[/tex] to it is [tex]GK=HK=\frac{68}{15}-\frac{18}{5}=\frac{14}{15}[/tex].

Solution:
The right triangles have equal acute angles, so [tex]\triangle ACD \sim \triangle CBD[/tex] with correspondence [tex]A \leftrightarrow C;\ C \leftrightarrow B;\ D \leftrightarrow D[/tex] and ratio [tex]\frac{AC}{CB}=\frac{b}{a}[/tex]. The side [tex]CD[/tex] of the first triangle corresponds to the side [tex]BD[/tex] of the second, so the medians [tex]AM[/tex] and [tex]CN[/tex] to these sides correspond.
Corresponding medians of similar triangles are in the similarity ratio, hence [tex]AM:CN=b:a[/tex].

Solution:
[tex]\triangle CMN \sim \triangle CAB[/tex] and [tex]\triangle AMP \sim \triangle ACB[/tex], so [tex]S_{CMN}=\left(\frac{1}{3}\right)^2S=\frac{S}{9}[/tex] and [tex]S_{AMP}=\left(\frac{2}{3}\right)^2S=\frac{4S}{9}[/tex].
[tex]MNBP[/tex] is a parallelogram ([tex]MN \parallel AB[/tex], [tex]MP \parallel BC[/tex]), so [tex]S_{MNBP}=S-\frac{S}{9}-\frac{4S}{9}=\frac{4S}{9}[/tex], and the diagonal [tex]NP[/tex] halves it: [tex]S_{MNP}=\frac{1}{2}S_{MNBP}=\frac{2S}{9}[/tex].

Solution:
Since [tex]AM \parallel CD[/tex], [tex]\triangle APM \sim \triangle CPD[/tex], so [tex]\frac{AP}{PC}=\frac{PM}{PD}=\frac{2}{3}[/tex].
[tex]S_{ACD}=\frac{15}{2}[/tex]. Triangles with a common vertex and bases on one line have areas proportional to the bases, so [tex]S_{APD}=\frac{2}{5}S_{ACD}=3[/tex] and [tex]S_{AMP}=\frac{2}{3}S_{APD}=2[/tex].

Solution:
The triangles have the common angle at [tex]A[/tex] and [tex]\angle ABM=\angle ACB[/tex], so [tex]\triangle ABM \sim \triangle ACB[/tex] and [tex]\frac{AB}{AC}=\frac{AM}{AB}[/tex].
Hence [tex]AB^2=AC \times AM=AC \times \frac{AC}{2}[/tex], i.e. [tex]AC^2=2AB^2[/tex].

Solution:
Let [tex]CD[/tex] be the diameter of the circumscribed circle through [tex]C[/tex]. Then [tex]\angle CBD=90^\circ[/tex], because it is inscribed in a semicircle, and [tex]\angle CAH=\angle CDB[/tex] (inscribed angles on the arc [tex]CB[/tex]), so [tex]\triangle CAH \sim \triangle CDB[/tex] and [tex]\frac{CA}{CD}=\frac{CH}{CB}[/tex].
Hence [tex]CD=\frac{CA \times CB}{CH}=\frac{12 \times 15}{10}=18[/tex], so [tex]R=9[/tex].

Solution:
[tex]BD=8[/tex]. Let [tex]AD=x[/tex] and place [tex]D(0,0),\ B(0,8),\ A(-x,0),\ C(21-x,0)[/tex]; then [tex]M\left(\frac{21-x}{2},4\right)[/tex]. The line [tex]AM[/tex] meets [tex]BD[/tex] at the height [tex]\frac{8x}{21+x}[/tex], so [tex]\frac{8x}{21+x}=\frac{10}{3}[/tex], which gives [tex]x=15[/tex] and [tex]DC=21-x=6[/tex].
Hence [tex]AB=\sqrt{15^2+8^2}=17;\ BC=\sqrt{6^2+8^2}=10[/tex].

Solution:
In [tex]\triangle AMC[/tex], the segment [tex]AO[/tex] bisects the angle at [tex]A[/tex], and [tex]AM=\frac{AB}{2}[/tex]. By the angle bisector theorem [tex]\frac{CO}{OM}=\frac{AC}{AM}=\frac{AC}{\frac{1}{2}AB}=\frac{3}{1}[/tex].