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Prove-that-if-A-B-and-C-are-the-midpoints-of-the-sides-BC-CA-and-AB-respectively-then-AA-BB-CC-lt-AB-BC-CA-

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Question Number 13888 by Tinkutara last updated on 24/May/17
Prove that if A′, B′ and C′ are the midpoints of the sides BC, CA and AB, respectively, then AA′ + BB′ + CC′ < AB + BC + CA
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{if}\:{A}',\:{B}'\:\mathrm{and}\:{C}'\:\mathrm{are}\:\mathrm{the} \\ $$$$\mathrm{midpoints}\:\mathrm{of}\:\mathrm{the}\:\mathrm{sides}\:{BC},\:{CA}\:\mathrm{and}\:{AB}, \\ $$$$\mathrm{respectively},\:\mathrm{then} \\ $$$${AA}'\:+\:{BB}'\:+\:{CC}'\:<\:{AB}\:+\:{BC}\:+\:{CA} \\ $$
Commented by Tinkutara last updated on 24/May/17
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17
AG+BG>AB⇒(2/3)(AA^′ +BB^′ )>AB BG+GC>BC⇒(2/3)(BB^′ +CC^′ )>BC AG+GC>AC⇒(2/3)(AA^′ +CC^′ )>AC ⇒1)AB+BC+CA<(4/3)(AA^′ +BB^′ +CC^′ ) 2)((∣AB−AC∣)/2)<AA^′ <((AB+AC)/2) 3)a(m_a ^2 +(a/2).(a/2))=b^2 .(a/2)+c^2 .(a/2)(Estewart′s teorem) ⇒m_a ^2 +(a^2 /4)=((b^2 +c^2 )/2)⇒m_a ^2 =((2(b^2 +c^2 )−a^2 )/4) =((2(b^2 +c^2 )−(b^2 +c^2 −2bc.cosA))/4)=(1/4)(b^2 +c^2 +2bc.cosA)= =(1/4)(b^2 +c^2 +2bc.cosA)<(1/4)(b^2 +c^2 +2bc)= =(1/4)(b+c)^2 ⇒m_a <(1/2)(b+c) similarly:⇒m_b <(1/2)(a+c),m_c <(1/2)(a+b) ⇒m_a +m_b +m_c <(1/2)(2a+2b+2c)=a+b+c ⇒m_a +m_b +m_c <a+b+c .■ 4)if: m_a =((b+c)/2)⇒((2(b^2 +c^2 )−a^2 )/4)=((b^2 +c^2 +2bc)/4) ⇒b^2 +c^2 −2bc=a^2 ⇒(b−c)^2 =a^2 ⇒b−c=a(impossible) or:b^2 +c^2 −2bc=b^2 +c^2 −2bc.cosA ⇒cosA=1⇒∡A=0 (impossible ) so: m_a ≰(1/2)(b+c) and: m_a <(1/2)(b+c) . 5)A=90^° ⇒4m_a ^2 =b^2 +c^2 =a^2 ⇒m_a =(a/2). A=120^° ⇒4m_a ^2 =b^2 +c^2 −bc A=60⇒ 4m_a ^2 =b^2 +c^2 +bc 6)A=90,B=60,C=30 ⇒4(m_a ^2 +m_b ^2 +m_c ^2 )=a^2 +a^2 +c^2 +ac+a^2 +b^2 +(√3)ab= =4a^2 +(√3)a(b+c). 7)A=90,b=c⇒Σm_a ^2 =4a^2 +2(√2)ab .^ 8)BC^2 =BG^2 +GC^2 −2BG.GC.cosBG^� C ∡BGC=ϕ⇒a^2 =(4/9)m_b ^2 +(4/9)m_c ^2 −2(2/3)m_b .(2/3)m_c cosϕ ⇒(9/4)a^2 =m_b ^2 +m_c ^2 −2m_b .m_c .cosϕ ϕ=90⇒(9/4)a^2 =((2(a^2 +c^2 )−b^2 )/4)+((2(a^2 +b^2 )−c^2 )/4)⇒ ⇒(if:m_b ⊥m_c )⇒(5a^2 =b^2 +c^2 ),(m_a =(3/2)a) 9)AB⊥AC⇒m_b ^2 =((2(a^2 +c^2 )−b^2 )/4)=((b^2 +2c^2 )/4) m_c ^2 =((2(a^2 +b^2 )−c^2 )/4)=((2b^2 +c^2 )/4) ⇒m_b ^2 +m_c ^2 =(3/4)(b^2 +c^2 )=(3/4)a^2 =3m_a ^2 ⇒ { ((b⊥c⇒ m_b ^2 +m_c ^2 =3m_a ^2 ,m_a =(1/2)a .)),((m_b ⊥m_c ⇒b^2 +c^2 =5a^2 ,m_a =(3/2)a .)) :}
$${AG}+{BG}>{AB}\Rightarrow\frac{\mathrm{2}}{\mathrm{3}}\left({AA}^{'} +{BB}^{'} \right)>{AB} \\ $$$${BG}+{GC}>{BC}\Rightarrow\frac{\mathrm{2}}{\mathrm{3}}\left({BB}^{'} +{CC}^{'} \right)>{BC} \\ $$$${AG}+{GC}>{AC}\Rightarrow\frac{\mathrm{2}}{\mathrm{3}}\left({AA}^{'} +{CC}^{'} \right)>{AC} \\ $$$$\left.\Rightarrow\mathrm{1}\right){AB}+{BC}+{CA}<\frac{\mathrm{4}}{\mathrm{3}}\left({AA}^{'} +{BB}^{'} +{CC}^{'} \right) \\ $$$$\left.\mathrm{2}\right)\frac{\mid{AB}−{AC}\mid}{\mathrm{2}}<{AA}^{'} <\frac{{AB}+{AC}}{\mathrm{2}} \\ $$$$\left.\mathrm{3}\right){a}\left({m}_{{a}} ^{\mathrm{2}} +\frac{{a}}{\mathrm{2}}.\frac{{a}}{\mathrm{2}}\right)={b}^{\mathrm{2}} .\frac{{a}}{\mathrm{2}}+{c}^{\mathrm{2}} .\frac{{a}}{\mathrm{2}}\left({Estewart}'{s}\:{teorem}\right) \\ $$$$\Rightarrow{m}_{{a}} ^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}\Rightarrow{m}_{{a}} ^{\mathrm{2}} =\frac{\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$=\frac{\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}.{cosA}\right)}{\mathrm{4}}=\frac{\mathrm{1}}{\mathrm{4}}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{bc}.{cosA}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{bc}.{cosA}\right)<\frac{\mathrm{1}}{\mathrm{4}}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{bc}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\left({b}+{c}\right)^{\mathrm{2}} \Rightarrow{m}_{{a}} <\frac{\mathrm{1}}{\mathrm{2}}\left({b}+{c}\right) \\ $$$${similarly}:\Rightarrow{m}_{{b}} <\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{c}\right),{m}_{{c}} <\frac{\mathrm{1}}{\mathrm{2}}\left({a}+{b}\right) \\ $$$$\Rightarrow{m}_{{a}} +{m}_{{b}} +{m}_{{c}} <\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2}{a}+\mathrm{2}{b}+\mathrm{2}{c}\right)={a}+{b}+{c} \\ $$$$\Rightarrow{m}_{{a}} +{m}_{{b}} +{m}_{{c}} <{a}+{b}+{c}\:\:\:.\blacksquare \\ $$$$\left.\mathrm{4}\right){if}:\:{m}_{{a}} =\frac{{b}+{c}}{\mathrm{2}}\Rightarrow\frac{\mathrm{2}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{bc}}{\mathrm{4}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}={a}^{\mathrm{2}} \Rightarrow\left({b}−{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} \Rightarrow{b}−{c}={a}\left({impossible}\right) \\ $$$${or}:{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{bc}.{cosA} \\ $$$$\Rightarrow{cosA}=\mathrm{1}\Rightarrow\measuredangle{A}=\mathrm{0}\:\left({impossible}\:\right) \\ $$$${so}:\:{m}_{{a}} \nleqslant\frac{\mathrm{1}}{\mathrm{2}}\left({b}+{c}\right)\:{and}:\:{m}_{{a}} <\frac{\mathrm{1}}{\mathrm{2}}\left({b}+{c}\right)\:. \\ $$$$\left.\mathrm{5}\right){A}=\mathrm{90}^{°} \Rightarrow\mathrm{4}{m}_{{a}} ^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}^{\mathrm{2}} \Rightarrow{m}_{{a}} =\frac{{a}}{\mathrm{2}}. \\ $$$$\:\:\:\:{A}=\mathrm{120}^{°} \Rightarrow\mathrm{4}{m}_{{a}} ^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{bc} \\ $$$$\:\:\:\:{A}=\mathrm{60}\Rightarrow\:\:\:\mathrm{4}{m}_{{a}} ^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{bc} \\ $$$$\left.\mathrm{6}\right){A}=\mathrm{90},{B}=\mathrm{60},{C}=\mathrm{30} \\ $$$$\Rightarrow\mathrm{4}\left({m}_{{a}} ^{\mathrm{2}} +{m}_{{b}} ^{\mathrm{2}} +{m}_{{c}} ^{\mathrm{2}} \right)={a}^{\mathrm{2}} +{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +{ac}+{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +\sqrt{\mathrm{3}}{ab}= \\ $$$$=\mathrm{4}{a}^{\mathrm{2}} +\sqrt{\mathrm{3}}{a}\left({b}+{c}\right). \\ $$$$\left.\mathrm{7}\right){A}=\mathrm{90},{b}={c}\Rightarrow\Sigma{m}_{{a}} ^{\mathrm{2}} =\mathrm{4}{a}^{\mathrm{2}} +\mathrm{2}\sqrt{\mathrm{2}}{ab}\:\overset{} {.} \\ $$$$\left.\mathrm{8}\right){BC}^{\mathrm{2}} ={BG}^{\mathrm{2}} +{GC}^{\mathrm{2}} −\mathrm{2}{BG}.{GC}.{cosB}\overset{} {{G}C} \\ $$$$\measuredangle{BGC}=\varphi\Rightarrow{a}^{\mathrm{2}} =\frac{\mathrm{4}}{\mathrm{9}}{m}_{{b}} ^{\mathrm{2}} +\frac{\mathrm{4}}{\mathrm{9}}{m}_{{c}} ^{\mathrm{2}} −\mathrm{2}\frac{\mathrm{2}}{\mathrm{3}}{m}_{{b}} .\frac{\mathrm{2}}{\mathrm{3}}{m}_{{c}} {cos}\varphi \\ $$$$\Rightarrow\frac{\mathrm{9}}{\mathrm{4}}{a}^{\mathrm{2}} ={m}_{{b}} ^{\mathrm{2}} +{m}_{{c}} ^{\mathrm{2}} −\mathrm{2}{m}_{{b}} .{m}_{{c}} .{cos}\varphi \\ $$$$\varphi=\mathrm{90}\Rightarrow\frac{\mathrm{9}}{\mathrm{4}}{a}^{\mathrm{2}} =\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{c}^{\mathrm{2}} }{\mathrm{4}}\Rightarrow \\ $$$$\Rightarrow\left({if}:{m}_{{b}} \bot{m}_{{c}} \right)\Rightarrow\left(\mathrm{5}{a}^{\mathrm{2}} ={b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right),\left({m}_{{a}} =\frac{\mathrm{3}}{\mathrm{2}}{a}\right) \\ $$$$\left.\mathrm{9}\right){AB}\bot{AC}\Rightarrow{m}_{{b}} ^{\mathrm{2}} =\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} }{\mathrm{4}}=\frac{{b}^{\mathrm{2}} +\mathrm{2}{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}_{{c}} ^{\mathrm{2}} =\frac{\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)−{c}^{\mathrm{2}} }{\mathrm{4}}=\frac{\mathrm{2}{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\Rightarrow{m}_{{b}} ^{\mathrm{2}} +{m}_{{c}} ^{\mathrm{2}} =\frac{\mathrm{3}}{\mathrm{4}}\left({b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)=\frac{\mathrm{3}}{\mathrm{4}}{a}^{\mathrm{2}} =\mathrm{3}{m}_{{a}} ^{\mathrm{2}} \\ $$$$\Rightarrow\begin{cases}{{b}\bot{c}\Rightarrow\:\:\:\:\:\:{m}_{{b}} ^{\mathrm{2}} +{m}_{{c}} ^{\mathrm{2}} =\mathrm{3}{m}_{{a}} ^{\mathrm{2}} ,{m}_{{a}} =\frac{\mathrm{1}}{\mathrm{2}}{a}\:.}\\{{m}_{{b}} \bot{m}_{{c}} \Rightarrow{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{5}{a}^{\mathrm{2}} ,{m}_{{a}} =\frac{\mathrm{3}}{\mathrm{2}}{a}\:.}\end{cases} \\ $$$$\:\:\:\:\: \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 25/May/17
in case of:∡AGB=∡BGC=∡AGC show that: 1) S_(ABC) =((√3)/(12)).(a^2 +b^2 +c^2 ) 2)(a/h_a )+(b/h_b )+(c/h_c )=2(√3) . (h_a =altitude from A^� to BC)
$${in}\:{case}\:{of}:\measuredangle{AGB}=\measuredangle{BGC}=\measuredangle{AGC} \\ $$$${show}\:{that}: \\ $$$$\left.\mathrm{1}\right)\:{S}_{{ABC}} =\frac{\sqrt{\mathrm{3}}}{\mathrm{12}}.\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right) \\ $$$$\left.\mathrm{2}\right)\frac{{a}}{{h}_{{a}} }+\frac{{b}}{{h}_{{b}} }+\frac{{c}}{{h}_{{c}} }=\mathrm{2}\sqrt{\mathrm{3}}\:\:.\:\:\left({h}_{{a}} ={altitude}\:{from}\:\overset{} {{A}}\:{to}\:{BC}\right) \\ $$
Answered by mrW1 last updated on 24/May/17
AA′=(1/2)(√(AB^2 +AC^2 +2×AB×AC×cos ∠A)) <(1/2)(√(AB^2 +AC^2 +2×AB×AC)) =(1/2)(√((AB+AC)^2 ))=(1/2)(AB+AC) AA′<(1/2)(AB+AC) BB′<(1/2)(BA+BC) CC′<(1/2)(CB+CA) ⇒AA′ + BB′ + CC′ < AB + BC + CA
$${AA}'=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} +\mathrm{2}×{AB}×{AC}×\mathrm{cos}\:\angle{A}} \\ $$$$<\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} +\mathrm{2}×{AB}×{AC}}\: \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\left({AB}+{AC}\right)^{\mathrm{2}} }=\frac{\mathrm{1}}{\mathrm{2}}\left({AB}+{AC}\right) \\ $$$$ \\ $$$${AA}'<\frac{\mathrm{1}}{\mathrm{2}}\left({AB}+{AC}\right) \\ $$$${BB}'<\frac{\mathrm{1}}{\mathrm{2}}\left({BA}+{BC}\right) \\ $$$${CC}'<\frac{\mathrm{1}}{\mathrm{2}}\left({CB}+{CA}\right) \\ $$$$\Rightarrow{AA}'\:+\:{BB}'\:+\:{CC}'\:<\:{AB}\:+\:{BC}\:+\:{CA} \\ $$
Commented by mrW1 last updated on 24/May/17
You can also prove like this: since BA′=A′C ⇒A′A′′=A′A′′′ AA′′<AC AA′′′<AB AA′=(1/2)(AA′′+AA′′′)<(1/2)(AC+AB)
$${You}\:{can}\:{also}\:{prove}\:{like}\:{this}: \\ $$$${since}\:{BA}'={A}'{C} \\ $$$$\Rightarrow{A}'{A}''={A}'{A}''' \\ $$$${AA}''<{AC} \\ $$$${AA}'''<{AB} \\ $$$${AA}'=\frac{\mathrm{1}}{\mathrm{2}}\left({AA}''+{AA}'''\right)<\frac{\mathrm{1}}{\mathrm{2}}\left({AC}+{AB}\right) \\ $$
Commented by mrW1 last updated on 24/May/17
Using law of cosines we get: AB^2 =BA′^2 +AA′^2 −2×BA′×AA′×cos ∠AA′B ...(i) AC^2 =CA′^2 +AA′^2 −2×CA′×AA′×cos ∠AA′C AC^2 =CA′^2 +AA′^2 −2×CA′×AA′×cos (180°−∠AA′B) AC^2 =CA′^2 +AA′^2 +2×CA′×AA′×cos ∠AA′B AC^2 =CA′^2 +AA′^2 +2×BA′×AA′×cos ∠AA′B ...(ii) (i)+(ii): AB^2 +AC^2 =BA′^2 +CA′^2 +2×AA′^2 AB^2 +AC^2 =(((BC)/2))^2 +(((BC)/2))^2 +2×AA′^2 AB^2 +AC^2 =((BC^2 )/2)+2×AA′^2 AA′^2 =(1/2)(AB^2 +AC^2 −((BC^2 )/2)) ...(iii) BC^2 =AB^2 +AC^2 −2×AB×AC×cos ∠A into (iii): AA′^2 =(1/4)(AB^2 +AC^2 +2×AB×AC×cos ∠A) ...(iii) ⇒AA′=(1/2)(√(AB^2 +AC^2 +2×AB×AC×cos ∠A))
$${Using}\:{law}\:{of}\:{cosines}\:{we}\:{get}: \\ $$$${AB}^{\mathrm{2}} ={BA}'^{\mathrm{2}} +{AA}'^{\mathrm{2}} −\mathrm{2}×{BA}'×{AA}'×\mathrm{cos}\:\angle{AA}'{B}\:\:\:\:…\left({i}\right) \\ $$$$ \\ $$$${AC}^{\mathrm{2}} ={CA}'^{\mathrm{2}} +{AA}'^{\mathrm{2}} −\mathrm{2}×{CA}'×{AA}'×\mathrm{cos}\:\angle{AA}'{C} \\ $$$${AC}^{\mathrm{2}} ={CA}'^{\mathrm{2}} +{AA}'^{\mathrm{2}} −\mathrm{2}×{CA}'×{AA}'×\mathrm{cos}\:\left(\mathrm{180}°−\angle{AA}'{B}\right) \\ $$$${AC}^{\mathrm{2}} ={CA}'^{\mathrm{2}} +{AA}'^{\mathrm{2}} +\mathrm{2}×{CA}'×{AA}'×\mathrm{cos}\:\angle{AA}'{B} \\ $$$${AC}^{\mathrm{2}} ={CA}'^{\mathrm{2}} +{AA}'^{\mathrm{2}} +\mathrm{2}×{BA}'×{AA}'×\mathrm{cos}\:\angle{AA}'{B}\:\:\:…\left({ii}\right) \\ $$$$ \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$${AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} ={BA}'^{\mathrm{2}} +{CA}'^{\mathrm{2}} +\mathrm{2}×{AA}'^{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} =\left(\frac{{BC}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{{BC}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{2}×{AA}'^{\mathrm{2}} \\ $$$${AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} =\frac{{BC}^{\mathrm{2}} }{\mathrm{2}}+\mathrm{2}×{AA}'^{\mathrm{2}} \\ $$$${AA}'^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}\left({AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} −\frac{{BC}^{\mathrm{2}} }{\mathrm{2}}\right)\:\:\:…\left({iii}\right) \\ $$$$ \\ $$$${BC}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} −\mathrm{2}×{AB}×{AC}×\mathrm{cos}\:\angle{A} \\ $$$${into}\:\left({iii}\right): \\ $$$${AA}'^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{4}}\left({AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} +\mathrm{2}×{AB}×{AC}×\mathrm{cos}\:\angle{A}\right)\:\:\:…\left({iii}\right) \\ $$$$\Rightarrow{AA}'=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} +\mathrm{2}×{AB}×{AC}×\mathrm{cos}\:\angle{A}} \\ $$
Commented by ajfour last updated on 24/May/17
how nice !
$${how}\:{nice}\:! \\ $$
Commented by mrW1 last updated on 24/May/17
Commented by ajfour last updated on 24/May/17
all tbe more better !
$${all}\:{tbe}\:{more}\:{better}\:!\: \\ $$
Commented by Rishabh#1 last updated on 24/May/17
How did u get AA′=(1/2)(√(AB^2 +AC^2 +2×AB×AC×cos ∠A))
$$\mathrm{How}\:\mathrm{did}\:\mathrm{u}\:\mathrm{get} \\ $$$${AA}'=\frac{\mathrm{1}}{\mathrm{2}}\sqrt{{AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} +\mathrm{2}×{AB}×{AC}×\mathrm{cos}\:\angle{A}} \\ $$
Commented by Tinkutara last updated on 25/May/17
Thanks.
$$\mathrm{Thanks}. \\ $$

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