Question Number 220546 by mr W last updated on 15/May/25
Commented by mr W last updated on 15/May/25
$${proof}\:{to}\:{Q}\mathrm{219318} \\ $$
Commented by mr W last updated on 15/May/25
![WLOG assume A(0, a) B(b, 0) C(c, 0) ∣BM∣=μ∣BC∣, ∣B′M′∣=μ∣B′C′∣ μ is any contant. μ=0.5 for M being midpoint of BC M(x_M , 0) x_M =b+μ(c−b)=(1−μ)b+μc ((x_(B′) −x_B )/a)=(y_(B′) /b)=((BB′)/(AB))=tan θ=λ ⇒x_(B′) =b+λa, y_(B′) =λb similarly ⇒x_(C′) =c+λa, y_(C′) =λc x_(M′) =(1−μ)x_(B′) +μx_(C′) =(1−μ)(b+λa)+μ(c+λa) =(1−μ)b+μc+λa=x_M +λa y_(M′) =(1−μ)y_(B′) +μy_(C′) =(1−μ)(λb)+μ(λc) =λ[(1−μ)b+μc]=λx_M ∣AM∣^2 =x_M ^2 +a^2 ∣AM′∣^2 =x_(M′) ^2 +(a−y_(M′) )^2 =(x_M +λa)^2 +(a−λx_M )^2 =(1+λ^2 )(x_M ^2 +a^2 ) ∣MM′∣^2 =(x_(M′) −x_M )^2 +y_(M′) ^2 =(x_M +λa−x_M )^2 +(λx_M )^2 =λ^2 (a^2 +x_M ^2 ) we see ∣AM′∣^2 =∣AM∣^2 +∣MM′∣^2 and ((∣MM′∣)/(∣AM∣))=(√((λ^2 (a^2 +x_M ^2 ))/(a^2 +x_M ^2 )))=λ=tan θ that means MM′⊥AM and ∠MAM′=θ](https://www.tinkutara.com/question/Q220548.png)
$${WLOG}\:{assume} \\ $$$${A}\left(\mathrm{0},\:{a}\right) \\ $$$${B}\left({b},\:\mathrm{0}\right) \\ $$$${C}\left({c},\:\mathrm{0}\right) \\ $$$$\mid{BM}\mid=\mu\mid{BC}\mid,\:\mid{B}'{M}'\mid=\mu\mid{B}'{C}'\mid \\ $$$$\mu\:{is}\:{any}\:{contant}.\: \\ $$$$\mu=\mathrm{0}.\mathrm{5}\:{for}\:{M}\:{being}\:{midpoint}\:{of}\:{BC} \\ $$$${M}\left({x}_{{M}} ,\:\mathrm{0}\right) \\ $$$${x}_{{M}} ={b}+\mu\left({c}−{b}\right)=\left(\mathrm{1}−\mu\right){b}+\mu{c} \\ $$$$\frac{{x}_{{B}'} −{x}_{{B}} }{{a}}=\frac{{y}_{{B}'} }{{b}}=\frac{{BB}'}{{AB}}=\mathrm{tan}\:\theta=\lambda \\ $$$$\Rightarrow{x}_{{B}'} ={b}+\lambda{a},\:{y}_{{B}'} =\lambda{b} \\ $$$${similarly} \\ $$$$\Rightarrow{x}_{{C}'} ={c}+\lambda{a},\:{y}_{{C}'} =\lambda{c} \\ $$$${x}_{{M}'} =\left(\mathrm{1}−\mu\right){x}_{{B}'} +\mu{x}_{{C}'} \\ $$$$\:\:\:\:\:\:\:=\left(\mathrm{1}−\mu\right)\left({b}+\lambda{a}\right)+\mu\left({c}+\lambda{a}\right) \\ $$$$\:\:\:\:\:\:\:=\left(\mathrm{1}−\mu\right){b}+\mu{c}+\lambda{a}={x}_{{M}} +\lambda{a} \\ $$$${y}_{{M}'} =\left(\mathrm{1}−\mu\right){y}_{{B}'} +\mu{y}_{{C}'} \\ $$$$\:\:\:\:\:\:\:=\left(\mathrm{1}−\mu\right)\left(\lambda{b}\right)+\mu\left(\lambda{c}\right) \\ $$$$\:\:\:\:\:\:\:=\lambda\left[\left(\mathrm{1}−\mu\right){b}+\mu{c}\right]=\lambda{x}_{{M}} \\ $$$$\mid{AM}\mid^{\mathrm{2}} ={x}_{{M}} ^{\mathrm{2}} +{a}^{\mathrm{2}} \\ $$$$\mid{AM}'\mid^{\mathrm{2}} ={x}_{{M}'} ^{\mathrm{2}} +\left({a}−{y}_{{M}'} \right)^{\mathrm{2}} =\left({x}_{{M}} +\lambda{a}\right)^{\mathrm{2}} +\left({a}−\lambda{x}_{{M}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\left(\mathrm{1}+\lambda^{\mathrm{2}} \right)\left({x}_{{M}} ^{\mathrm{2}} +{a}^{\mathrm{2}} \right) \\ $$$$\mid{MM}'\mid^{\mathrm{2}} =\left({x}_{{M}'} −{x}_{{M}} \right)^{\mathrm{2}} +{y}_{{M}'} ^{\mathrm{2}} =\left({x}_{{M}} +\lambda{a}−{x}_{{M}} \right)^{\mathrm{2}} +\left(\lambda{x}_{{M}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\lambda^{\mathrm{2}} \left({a}^{\mathrm{2}} +{x}_{{M}} ^{\mathrm{2}} \right) \\ $$$${we}\:{see}\: \\ $$$$\mid{AM}'\mid^{\mathrm{2}} =\mid{AM}\mid^{\mathrm{2}} +\mid{MM}'\mid^{\mathrm{2}} \:{and} \\ $$$$\frac{\mid{MM}'\mid}{\mid{AM}\mid}=\sqrt{\frac{\lambda^{\mathrm{2}} \left({a}^{\mathrm{2}} +{x}_{{M}} ^{\mathrm{2}} \right)}{{a}^{\mathrm{2}} +{x}_{{M}} ^{\mathrm{2}} }}=\lambda=\mathrm{tan}\:\theta \\ $$$${that}\:{means}\: \\ $$$${MM}'\bot{AM}\:{and}\:\angle{MAM}'=\theta \\ $$
Commented by ea last updated on 15/May/25
This is perfect ,Prof��
Very impressive ��