\mathsf{a = AB = 6, b =AC = 7 :c = BC = 5
\\ c^2 = a^2+b^2-2abcos\theta \implies 5^2 = 6^2+7^2-2.6.7.cos\theta \\\therefore cos\theta =\frac{5}{7}\\
x=acos\theta = 6.\frac{5}{7} \therefore \boxed{x =...

\mathsf{

S = \sqrt{p(p-a)(p-b)(p-c)}\\
p = \frac{13+14+15}{2} =21\\
\therefore S =\sqrt{21(21-13)(21-14)(21-15)} \implies S = 84\\
S= \frac{b.h}{2} \implies 84 = \frac{14.h}{2} \therefore \boxed{h...

\mathsf{
AF = x \implies BM=BN=5-x \implies CN = CF = 6-(5-x)=1+x \\
\therefore x+x+2(5-x)+2(1+x) = 5+6+7 \implies 2x= 6 \\
\therefore x = AC = 3\\

T.Stweart: BF^2.7 = 6^2.3+5^2.4-3.4.7 \implies...

\mathsf{
\triangle ABQ: 13^2 = h^2+AQ^2 \implies h^2 = 169-AQ^2\\
\triangle CND: 15^2 = h^2+(18-4-AQ)^2 \implies h^2 =225 - 196+2AQ-AQ^2\\
Igualando: 169-AQ^2 =29-28A+AQ^2 \implies AQ = 5 \implies AN...

\mathsf{AB=BC=CD=AD=l =2+1 = 3\\
T.Stweart:\\
\triangle ABC - F:
4^2.3= 3^2.1+BD^2.2-2.1.3 \implies BD^2 =\frac{45}{2}\\
CD^2+BD^2 = 4l^2 \implies CD^2 =36-\frac{45}{2}=\frac{27}{2} \\
\triangle...

\mathsf{
\triangle AMC: h^2 = 25 - (2-AH)^2 = 21 +4AH - AH^2\\
\triangle HBD: h^2 = 49-(4+AH)^2 = 33-8AH-AH^2\\
Igualando~h^2: 12 = 12AH \implies AH =1\\
\therefore h^2 = 21+4(1)-1 = 24 \implies h =...

\mathsf{

p_{ABC} =\frac{9+8+7}{2}=12\\
S_{ABC} = \sqrt{12(12-7)(12-8)(12-9)}=\sqrt{720} \\
\sqrt{720} = \frac{8.BD}{2} \implies BD = \frac{\sqrt{720}}{4}=3\sqrt5\\
\triangle ABD: BD^2+AD^2 = 7^2...

\mathsf{
p_{ABC} = \frac{5+6+ 7}{2} = 9\\
S_{ABC} = \sqrt{9(9-5)(9-6)(9-7)} = \sqrt{216}=6\sqrt6\\
\therefore 6\sqrt6 = \frac{BD.6}{2} \implies BD = 2\sqrt6\\
\triangle ABD:5^2 = BD^2+AD^2 \implies...

\mathsf{
\triangle ABN \cong \triangle CDM \implies BN \cong D=h\\
\triangle DMC:h^2 = 5^2 -(10-AM)^2=25-100+20AM-AM^2\\
\triangle AMD:h^2 = 64-AM^2\\
Igualando~ h^2: AM=\frac{139}{20}\\
MN=10-2AN...

\mathsf{AD=BC=12\\

\triangle ABH: BH^2 = 6^2 - AH^2\\
\triangle BHD: BH^2 = 8^2 - ((12-AH)^2\\
Igualando ~BH^2: 36 - AH^2 = 64-144+24AH-AH^2 \implies 24AH =116 \\
\therefore AH = \frac{29}{6}...

△CMD : CM=6:MD=7: DC=5, \\
h = BF\\
x = CK\\
\triangle MCK: h^2=6^2−x^2\\
\triangle MKD: h^2=7^2−(5−x)^2\\
Igualando ~h^2: 36 - x^2 = 49-25+10x-x^2 \implies 12 = 10x\\
\therefore x=\frac{6}{5}...

a^2 = b^2+c^2-2bccos \measuredangle BAC\\
a^2 = b^2+c^2+ \sqrt3bc \\
igualando ~ a^2: \cancel{b^2+c^2}-2\cancel{bc}cos \measuredangle BAC = \cancel{b^2+c^2}+\sqrt3\cancel{bc} \\
\therefore...

\mathsf{

\triangle CND: h^2 = 49-(4+x)^2\\
\triangle ABM: h^2 = 25-x^2\\
Igualando ~h^2 : 49 - 16-8x-x^2 = 25-x^2 \implies x = 1\\
\therefore h^2 = 25-1 = 24\\
\triangle ACN: AC^2 = 5^2+h^2 = 25+24...

\mathsf{
\triangle CND: 7^2 = (5-AM)^2 + h^2 \implies h^2 = 24+10AM-AM^2\\
\triangle ABN: h^2 = 25 - AM^2\\
Igualando ~h^2: 24+10AM-AM^2 = 25-AM^2 \implies \boxed{AM = 0,1}

}

Para mim faltam dados pois o valor de x não é uma constante.

Temos uma arbelos onde x é dado por \frac{(R.r)(R+r)}{R^2+r^2+Rr}

\mathsf{\triangle AHC \sim \triangle QHP\\
\triangle AQH \sim \triangle ABP
\therefore :\frac{BP}{25}=\frac{QH}{AH}=\frac{QP}{35}⟹QP=BP⋅\frac{7}{5}\\
BP^2+PA^2=25^2\\
(30−BP)^2+PA^2=35^2\\...

a,b,c são os lados postos aos vértices correspondentes
Portanto d = a − c = 8.\\

\mathsf{
BM^2=\frac{a^2+c^2}{2}−\frac{b^2}{4}\\
\triangle BDM \sim \triangle GFM\\
\text{Dado que o baricentro está...

Lo intenté de esta manera pero falta algo.
Medianas: x:y :z\\
a=BC: b = AC : c = AB\\
T.Medianas:\\
2x^2+\frac{b^2}{2} = a^2+c^2\\
2y^2+\frac{c^2}{2} = a^2+b^2\\
2z^2+\frac{a^2}{2} = b^2+c^2\\...

O triângulo que possui o circuncentro e o incentro coincidentes é o equilátero.
Por propriedade a distância do incentro a circunferência é a mesma da circunferência ao exincentro

Sendo R essa...

\mathsf{}

\triangle ABD: h^2 = 20^2 - AD^2 = 400-(11+DT)^2\\
\triangle CTD: h^2 = 13^2-DT^2\\
Igualando~h^2 : 110 =22x \implies x= 5\\
\therefore \boxed{MN = \frac{11+16}{2} = 13,5}

\mathsf{

\triangle OPM: (20-x)^2 = h^2+ 10^2 \implies h^2 = 400-40x+x^2-100 = 300-40x+x^2\\
\triangle PNM: (x+5)^2 = h^2+5^2 \implies h^2 = x^2+10x+25-55 = x^2+10x\\
Igualando ~h^2:...

\mathsf{CD⊥OA\\
OC=R−2r+x\\
OD=R−x\\
DE=r+x\\
△OCD_{ret} \\
△CED_{ret}\\
DE^2=EC^2+CD^2 \implies (r+x)^2=(r−x)^2+CD^2
\therefore CD^2=4rx\\
OD^2=OC^2+CD^2 \implies (R−x)^2=(R+x−2r)^2+4rx\\...

\triangle ABP: PD^2 = l^2+PA^2 - 2lPA\cos \theta(I)\\
\triangle PCB: PC^2 = l^2+PB^2 - 2lPB\cos \theta(II)\\
(I)+(II)PC^2+PD^2 = 2l^2+PA^2+PB^2-2l\cos \theta(PA+PB)\\
\underbrace{PC^2+PD^2} =2l^2+...

\mathsf{

PC=QP=y\\
PN=\frac{QD}{2}=x\\

\therefore \triangle MCP: MC^2=y^2+(2−x)^2\\

\triangle CQD: CD^2=(2y)^2+(2x)^2 \\
\therefore CD^2−2MC^2=4y^2+4x^2-2y^2-8+8x-2x^2 \implies...

\mathsf{
\triangle ABH: 24^2 = (13k)^2 - AH^2\\\
\triangle AHC: 24^2 = (37k)^2-(40k-AH)^2\\
Igualando: 169k^2 - AH^2 = 1369k^2-1600k^2+80kAH-AH^2\\
\therefore AH = 5k\\
\triangle ABH: 24^2 = (13k)^2...

\mathsf{x=BM=MQ\\
△CDM :(x+20)^2=20^2+(20−x)^2⟹x=5 m\\
\therefore DM=20+5=25 m: MC=20−5=15 m\\
FN = r=\frac{15⋅20}{15+20+25}=5 m\\
MN=20−(5+5)=10 m\\
cos\measuredangle...

Há algum erro no enunciado:

Distância ente baricentro (G) e o circuncentro (O) é dado por
OG = \sqrt{R^2-\frac{a^2+b^2+c^2}{9}}\\
\therefore OG = \sqrt{{(\sqrt{10}^2})-\frac{9}{9}}=\sqrt{10-1} =...

]$O'$ is the center of the smaller circle.
\mathsf {

\triangle IO'O:(r-x)^2+(IO')^2=(r+x)^2\implies IO'^2=r^2+2rx+x^2 -r^2+2rx-x^2 =4rx\\
\therefore IO'={\sqrt{4rx}}\\
\triangle...

Teorema de Stewart no i -ésimo segmento:

b^2 i \frac an + c^2 \frac{n-i}na = a( x_i^2 + \frac{a^2}{n^2}i(n-i))

Apliquemos um somatório dos dois lados: \sum_{i=1}^{n-1} dos dois lados:

b^2a...