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Question-199374

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Question Number 199374 by cortano12 last updated on 02/Nov/23
Commented by cortano12 last updated on 02/Nov/23
prove that B
$$\mathrm{prove}\:\mathrm{that}\:\:\cancel{\boldsymbol{{B}}} \\ $$
Commented by MathematicalUser2357 last updated on 03/Nov/23
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Commented by ajfour last updated on 03/Nov/23
consider a ⊥ dropped from A on BC . AH=h, BM=CM=k , MH=b AB^( 2) +AC^( 2) =2h^2 +(k+b)^2 +(k−b)^2 = 2h^2 +2k^2 +2b^2 now h^2 +b^2 =AM^( 2) , k^2 =BM^2 =CM^( 2) hence AB^( 2) +AC^( 2) =2AM^( 2) +2BM^( 2)
$${consider}\:{a}\:\bot\:{dropped}\:{from}\:{A}\:{on} \\ $$$${BC}\:.\:\:{AH}={h},\:\:{BM}={CM}={k}\:\:,\: \\ $$$${MH}={b} \\ $$$${AB}^{\:\mathrm{2}} +{AC}^{\:\mathrm{2}} =\mathrm{2}{h}^{\mathrm{2}} +\left({k}+{b}\right)^{\mathrm{2}} +\left({k}−{b}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{2}{h}^{\mathrm{2}} +\mathrm{2}{k}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} \\ $$$${now}\:\:{h}^{\mathrm{2}} +{b}^{\mathrm{2}} ={AM}^{\:\mathrm{2}} ,\:{k}^{\mathrm{2}} ={BM}^{\mathrm{2}} ={CM}^{\:\mathrm{2}} \\ $$$${hence}\:\:\:{AB}^{\:\mathrm{2}} +{AC}^{\:\mathrm{2}} =\mathrm{2}{AM}^{\:\mathrm{2}} +\mathrm{2}{BM}^{\:\mathrm{2}} \\ $$
Answered by AST last updated on 02/Nov/23
WLOG,translate M to origin,rotate BMC such that it coincides with the real axis Then m=0,b=−c (b−a)(b^− −a^− )+(c−a)(c^− −a^− )=2(m−a)(m^− −a^− ) +2(b−m)(b^− −m^− ) ⇔∣b∣^2 +∣a∣^2 −ab^− −a^− b+∣c∣^2 +∣a∣^2 −ac^− −ca^− =2∣m∣^2 +2∣a∣^2 −2am^− −2a^− m+2∣b∣^2 +2∣m∣^2 −2mb^− −2m^− b ⇔2∣b∣^2 +2∣a∣^2 −a(b^− +c^− )−a^− (b+c)=2∣a∣^2 +2∣b∣^2 ⇔0=0; Hence initial eqn must be true.
$${WLOG},{translate}\:{M}\:{to}\:{origin},{rotate}\:{BMC} \\ $$$${such}\:{that}\:{it}\:{coincides}\:{with}\:{the}\:{real}\:{axis} \\ $$$${Then}\:{m}=\mathrm{0},{b}=−{c} \\ $$$$\left({b}−{a}\right)\left(\overset{−} {{b}}−\overset{−} {{a}}\right)+\left({c}−{a}\right)\left(\overset{−} {{c}}−\overset{−} {{a}}\right)=\mathrm{2}\left({m}−{a}\right)\left(\overset{−} {{m}}−\overset{−} {{a}}\right) \\ $$$$+\mathrm{2}\left({b}−{m}\right)\left(\overset{−} {{b}}−\overset{−} {{m}}\right) \\ $$$$\Leftrightarrow\mid{b}\mid^{\mathrm{2}} +\mid{a}\mid^{\mathrm{2}} −{a}\overset{−} {{b}}−\overset{−} {{a}b}+\mid{c}\mid^{\mathrm{2}} +\mid{a}\mid^{\mathrm{2}} −{a}\overset{−} {{c}}−{c}\overset{−} {{a}} \\ $$$$=\mathrm{2}\mid{m}\mid^{\mathrm{2}} +\mathrm{2}\mid{a}\mid^{\mathrm{2}} −\mathrm{2}{a}\overset{−} {{m}}−\mathrm{2}\overset{−} {{a}m}+\mathrm{2}\mid{b}\mid^{\mathrm{2}} +\mathrm{2}\mid{m}\mid^{\mathrm{2}} −\mathrm{2}{m}\overset{−} {{b}} \\ $$$$−\mathrm{2}\overset{−} {{m}b} \\ $$$$\Leftrightarrow\mathrm{2}\mid{b}\mid^{\mathrm{2}} +\mathrm{2}\mid{a}\mid^{\mathrm{2}} −{a}\left(\overset{−} {{b}}+\overset{−} {{c}}\right)−\overset{−} {{a}}\left({b}+{c}\right)=\mathrm{2}\mid{a}\mid^{\mathrm{2}} +\mathrm{2}\mid{b}\mid^{\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{0}=\mathrm{0};\:{Hence}\:{initial}\:{eqn}\:{must}\:{be}\:{true}. \\ $$
Answered by som(math1967) last updated on 02/Nov/23
Answered by mr W last updated on 02/Nov/23
BM=CM cos ∠BMA=((AM^2 +BM^2 −AB^2 )/(2×AM×BM)) cos ∠AMC=((AM^2 +CM^2 −AC^2 )/(2×AM×CM))=−cos ∠BMA ((AM^2 +CM^2 −AC^2 )/(2×AM×CM))=−((AM^2 +BM^2 −AB^2 )/(2×AM×BM)) AM^2 +CM^2 −AC^2 =−(AM^2 +BM^2 −AB^2 ) ⇒2 AM^2 +2 BM^2 =AB^2 +AC^2
$${BM}={CM} \\ $$$$\mathrm{cos}\:\angle{BMA}=\frac{{AM}^{\mathrm{2}} +{BM}^{\mathrm{2}} −{AB}^{\mathrm{2}} }{\mathrm{2}×{AM}×{BM}} \\ $$$$\mathrm{cos}\:\angle{AMC}=\frac{{AM}^{\mathrm{2}} +{CM}^{\mathrm{2}} −{AC}^{\mathrm{2}} }{\mathrm{2}×{AM}×{CM}}=−\mathrm{cos}\:\angle{BMA} \\ $$$$\frac{{AM}^{\mathrm{2}} +{CM}^{\mathrm{2}} −{AC}^{\mathrm{2}} }{\mathrm{2}×{AM}×{CM}}=−\frac{{AM}^{\mathrm{2}} +{BM}^{\mathrm{2}} −{AB}^{\mathrm{2}} }{\mathrm{2}×{AM}×{BM}} \\ $$$${AM}^{\mathrm{2}} +{CM}^{\mathrm{2}} −{AC}^{\mathrm{2}} =−\left({AM}^{\mathrm{2}} +{BM}^{\mathrm{2}} −{AB}^{\mathrm{2}} \right) \\ $$$$\Rightarrow\mathrm{2}\:{AM}^{\mathrm{2}} +\mathrm{2}\:{BM}^{\mathrm{2}} ={AB}^{\mathrm{2}} +{AC}^{\mathrm{2}} \\ $$

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