Similar Triangles: Very Difficult Problems with Solutions

Solution:
Draw the bisector [tex]AL[/tex] of the angle at [tex]A[/tex]. Then [tex]\angle LAB=\angle ABL[/tex] and [tex]\angle CAL=\angle ABC[/tex], so [tex]AL=LB[/tex]. The triangles also share the angle at [tex]C[/tex], so [tex]\triangle CAL \sim \triangle CBA[/tex] and [tex]\frac{CA}{CB}=\frac{CL}{CA}=\frac{AL}{BA}[/tex].
So [tex]CL=\frac{CA^2}{CB}=\frac{64}{12}=\frac{16}{3}[/tex], [tex]AL=LB=12-\frac{16}{3}=\frac{20}{3}[/tex], and [tex]AB=\frac{AL \times CB}{CA}=\frac{20}{3} \times \frac{12}{8}=10[/tex].

Solution:
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], hence [tex]\frac{HD}{AD}=\frac{BD}{CD}[/tex], i.e. [tex]HD=\frac{AD \times DB}{CD}[/tex].
Therefore [tex]HD=\frac{CD}{2}[/tex] holds exactly when [tex]\frac{AD \times DB}{CD}=\frac{CD}{2}[/tex], i.e. exactly when [tex]CD^2=2AD \times DB[/tex].

Solution:
Let [tex]H[/tex] be the foot of the perpendicular from [tex]C[/tex] to [tex]AB[/tex], and let [tex]p[/tex] be the signed length of [tex]MH[/tex], positive in the direction of [tex]\overrightarrow{AB}[/tex]. By the Pythagorean theorem in the right triangles with vertex [tex]H[/tex]: [tex]AC^2=CM^2+AM^2+2AM \times p[/tex] and [tex]BC^2=CM^2+BM^2-2BM \times p[/tex].
Multiplying the first equality by [tex]BM[/tex], the second by [tex]AM[/tex] and adding, we get [tex]BM \times AC^2+AM \times BC^2=(AM+BM)CM^2+AM \times BM(AM+BM)[/tex]. Since [tex]AM+BM=AB[/tex], this is the required equality.

Solution:
Let [tex]O[/tex] be the intersection point of [tex]AM[/tex] and [tex]BL[/tex]. In [tex]\triangle ABM[/tex] the bisector from [tex]B[/tex] is also an altitude, so the triangle is isosceles: [tex]AB=BM[/tex], [tex]AO=OM=2[/tex] and [tex]BC=2AB[/tex]. By the angle bisector theorem [tex]\frac{AL}{LC}=\frac{AB}{BC}=\frac{1}{2}[/tex].
With [tex]\vec{BA}=\vec a;\ \vec{BM}=\vec m;\ |\vec a|=|\vec m|=s[/tex]: [tex]\vec{BO}=\frac{\vec a+\vec m}{2};\ \vec{BL}=\frac{2\vec a+2\vec m}{3}=\frac{4}{3}\vec{BO}[/tex], so [tex]BO=3[/tex]; the right [tex]\triangle ABO[/tex] gives [tex]s^2=BO^2+AO^2=13[/tex], hence [tex]AB=\sqrt{13};\ BC=2\sqrt{13}[/tex].
Also [tex]\vec a \cdot \vec m=s^2-2AO^2=5[/tex], so [tex]AC^2=|2\vec m-\vec a|^2=5s^2-4\vec a \cdot \vec m[/tex] gives [tex]AC^2=65-20=45;\ AC=3\sqrt{5}[/tex], and [tex]P=3\sqrt{13}+3\sqrt{5}[/tex].

Solution:
With the usual notation [tex]a=4;\ b=5;\ c=6[/tex], the angle bisector theorem gives [tex]AB_1=\frac{bc}{a+c};\ AC_1=\frac{bc}{a+b}[/tex], so for the corner triangle at [tex]A[/tex]: [tex]\frac{S_{AB_1C_1}}{S}=\frac{AB_1 \times AC_1}{AC \times AB}=\frac{bc}{(a+b)(a+c)}[/tex]. Numerically, and similarly for the other two corners: [tex]\frac{S_{AB_1C_1}}{S}=\frac{30}{90}=\frac{1}{3};\ \frac{S_{BC_1A_1}}{S}=\frac{24}{99}=\frac{8}{33};\ \frac{S_{CA_1B_1}}{S}=\frac{20}{110}=\frac{2}{11}[/tex].
Hence [tex]\frac{S_{A_1B_1C_1}}{S}=1-\frac{1}{3}-\frac{8}{33}-\frac{2}{11}=\frac{8}{33}[/tex], i.e. [tex]\frac{S_{ABC}}{S_{A_1B_1C_1}}=\frac{33}{8}[/tex].

Solution:
Let [tex]M[/tex] be the midpoint of [tex]BC[/tex] and [tex]H'[/tex] the point on the extension of [tex]OG[/tex] beyond [tex]G[/tex] with [tex]GH'=2OG[/tex]. The centroid divides the median [tex]AM[/tex] so that [tex]\frac{GA}{GM}=\frac{GH'}{GO}=2[/tex], and [tex]\angle AGH'=\angle MGO[/tex] (vertical angles), so [tex]\triangle AGH' \sim \triangle MGO[/tex].
Hence [tex]AH' \parallel OM[/tex]; since [tex]OM \perp BC[/tex], also [tex]AH' \perp BC[/tex], i.e. [tex]H'[/tex] lies on the altitude from [tex]A[/tex]. In the same way [tex]H'[/tex] lies on the altitude from [tex]B[/tex], so [tex]H'=H[/tex] and the points [tex]O;\ G;\ H[/tex] are collinear.

Solution:
[tex]CD[/tex] is perpendicular to [tex]PK[/tex], so it is an altitude of [tex]\triangle PCK[/tex] and [tex]H[/tex] lies on it; also [tex]PH \perp CK[/tex]. Both angles in [tex]\angle DPH=\angle DCK[/tex] complement the angle at [tex]K[/tex] to a right angle, so [tex]\triangle PDH \sim \triangle CDK[/tex] and [tex]\frac{HD}{PD}=\frac{DK}{CD}[/tex].
Since [tex]CD^2=AD \times BD[/tex], [tex]HD=\frac{PD \times DK}{CD}=\frac{AD \times BD}{4CD}=\frac{CD}{4}[/tex], so [tex]CH:HD=3:1[/tex].

Solution:
Let [tex]H[/tex] be the foot of the altitude from [tex]C[/tex], [tex]CH=h[/tex], [tex]AB=s[/tex] and [tex]AH=x;\ HB=y[/tex]. Since [tex]CH \parallel NA[/tex] (both are perpendicular to [tex]AB[/tex]), [tex]\frac{HQ}{QA}=\frac{CH}{NA}=\frac{h}{s}[/tex], and similarly from [tex]CH \parallel MB[/tex]: [tex]QA=\frac{xs}{h+s};\ BP=\frac{ys}{h+s}[/tex].
Since [tex]QP \parallel NM[/tex], the triangles with vertex [tex]C[/tex] are similar and [tex]\frac{QP}{NM}=\frac{h}{h+s};\ QP=\frac{hs}{h+s}[/tex]. In a right triangle [tex]xy=h^2[/tex], so [tex]AQ \times BP=\frac{xys^2}{(h+s)^2}=\frac{h^2s^2}{(h+s)^2}=QP^2[/tex].

Solution:
From [tex]\triangle MNB[/tex]: [tex]\angle NMB=180^\circ-\angle B-\varphi[/tex]. The angles at [tex]M[/tex] on the line [tex]AB[/tex] add up to a straight angle, so [tex]\angle PMN=180^\circ-\varphi-\angle NMB=\angle B[/tex].
In the same way [tex]\angle MNP=\angle C[/tex] and [tex]\angle NPM=\angle A[/tex], so the triangles have equal angles and [tex]\triangle ABC \sim \triangle PMN[/tex].

Solution:
[tex]\triangle ADC \sim \triangle CDB[/tex] with ratio [tex]\frac{AC}{CB}[/tex], and under this similarity the incenter [tex]O_1[/tex] corresponds to the incenter [tex]O_2[/tex], so [tex]\frac{DO_1}{DO_2}=\frac{AC}{CB}[/tex].
[tex]O_1[/tex] and [tex]O_2[/tex] lie on the bisectors of the right angles at [tex]D[/tex], so [tex]\angle O_1DO_2=45^\circ+45^\circ=90^\circ[/tex]. Since also [tex]\angle ACB=90^\circ[/tex], the two right triangles have proportional legs, hence [tex]\triangle O_1DO_2 \sim \triangle ACB[/tex].

Solution:
Let [tex]K[/tex] be the point of [tex]BC[/tex] with [tex]MK \parallel AS[/tex]. In [tex]\triangle ABS[/tex], [tex]M[/tex] is the midpoint of [tex]AB[/tex] and [tex]MK \parallel AS[/tex], so [tex]BK=KS[/tex]. In [tex]\triangle CMK[/tex], [tex]\frac{CS}{SK}=\frac{CP}{PM}=\frac{1}{2}[/tex], so [tex]CS=x;\ SK=KB=2x;\ a=5x;\ CS=\frac{a}{5}[/tex].
In the same way, since [tex]CQ:QM=2:1[/tex], we get [tex]CT=\frac{a}{2}[/tex]. Hence [tex]ST=\frac{a}{2}-\frac{a}{5}=\frac{3a}{10}[/tex].

Solution:
From the right triangles [tex]BP=AB\cos B;\ BQ=BC\cos B[/tex], so [tex]\triangle BPQ \sim \triangle BAC[/tex] with ratio [tex]k=\cos B[/tex]. The areas give [tex]k^2=\frac{6}{54}=\frac{1}{9}[/tex], so [tex]\cos B=\frac{1}{3};\ \sin B=\frac{2\sqrt{2}}{3}[/tex].
Then [tex]AC=\frac{PQ}{\cos B}=18\sqrt{2}[/tex] and by the law of sines [tex]R=\frac{AC}{2\sin B}=\frac{18\sqrt{2}}{\frac{4\sqrt{2}}{3}}=\frac{27}{2}[/tex].

Solution:
At [tex]C[/tex] we have [tex]CA_1=CA\cos C;\ CB_1=CB\cos C[/tex], so [tex]\triangle A_1B_1C \sim \triangle ABC[/tex] and [tex]\angle CA_1B_1=\angle CAB[/tex]; in the same way [tex]\angle BA_1C_1=\angle BAC[/tex].
Since [tex]AA_1 \perp BC[/tex], [tex]\angle B_1A_1A=90^\circ-\angle A=\angle C_1A_1A[/tex], so [tex]AA_1[/tex] bisects the angle of [tex]\triangle A_1B_1C_1[/tex] at [tex]A_1[/tex]. The other two altitudes are treated in the same way.

Solution:
Extend the legs to meet at [tex]E[/tex]. Then [tex]\triangle EDC \sim \triangle EMN \sim \triangle EAB[/tex], and the areas of similar triangles are proportional to the squares of corresponding sides: [tex]S_{EDC}=36t;\ S_{EMN}=x^2t;\ S_{EAB}=100t[/tex].
The two trapezoids have equal areas: [tex]x^2t-36t=100t-x^2t[/tex], so [tex]x^2=68[/tex] and [tex]x=2\sqrt{17}[/tex].