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Question Number 178837 by ARUNG_Brandon_MBU last updated on 22/Oct/22

Answered by HeferH last updated on 22/Oct/22

2(√3) ?

$$\mathrm{2}\sqrt{\mathrm{3}}\:? \\ $$

Commented by ARUNG_Brandon_MBU last updated on 22/Oct/22

It's among the options.

Commented by HeferH last updated on 22/Oct/22

Then the answer is 2(√3), I did it with the bisector theorem and pythagoras, you′ll get a cuadratic equation, we discard one of the solutions ( 4(√3) ) because it wouldn′t satisfy the other equation.

$${Then}\:{the}\:{answer}\:{is}\:\mathrm{2}\sqrt{\mathrm{3}},\:{I}\:{did}\:{it}\:{with}\:{the}\: \\ $$$${bisector}\:{theorem}\:{and}\:{pythagoras},\:{you}'{ll} \\ $$$$\:{get}\:\:{a}\:{cuadratic}\:{equation},\:{we}\:{discard}\:{one} \\ $$$$\:{of}\:{the}\:{solutions}\:\left(\:\mathrm{4}\sqrt{\mathrm{3}}\:\right)\:{because}\:{it}\:{wouldn}'{t} \\ $$$$\:{satisfy}\:{the}\:{other}\:{equation}. \\ $$$$\: \\ $$

Commented by HeferH last updated on 22/Oct/22

Its AD bisector?

$${Its}\:{AD}\:{bisector}? \\ $$

Commented by ARUNG_Brandon_MBU last updated on 22/Oct/22

For sure since we have BAC =60° and then two dots there representing equal angles 30°+30°

$$\mathrm{For}\:\mathrm{sure}\:\mathrm{since}\:\mathrm{we}\:\mathrm{have}\:\mathrm{BAC}\:=\mathrm{60}°\:\mathrm{and}\:\mathrm{then}\:\mathrm{two} \\ $$$$\mathrm{dots}\:\mathrm{there}\:\mathrm{representing}\:\mathrm{equal}\:\mathrm{angles}\:\mathrm{30}°+\mathrm{30}° \\ $$

Commented by HeferH last updated on 22/Oct/22

Commented by ARUNG_Brandon_MBU last updated on 22/Oct/22

Thank you !

Commented by HeferH last updated on 22/Oct/22

let AB = a, AC= c , AE = b 1. Since AD is bisector: (a/(3(√3))) = (c/( (√3))) a(√3) = 3(√3) c a=3c=AB 2. △AFC has a hipotenuse of c, then AF=(c/2) ; FC= ((c(√3))/2) 3. FB= AB−AF ⇒ FB= ((5c)/2) FB^2 + FC^2 = (4(√3))^2 ((25c^2 )/4) + ((3c^2 )/4)= 48 c= (√((48∙4)/(28))) = ((4(√3))/( (√7))) = ((4(√(21)))/7) 4. △AHC ≅ △AFC (Angle; side; Angle) AH = (c/2); CH= ((c(√3))/2) 5. AC is also bisector ((3c)/(4(√3))) = (b/x) b = ((3x(√7))/7) 6. HE = AE − AH = b−(c/2) CH^2 + HE^2 =x^2 ((3c^2 )/4) + (b−(c/2))^2 =x^2 (3/4)∙ (((4(√(21)))/7))^2 + (((3x(√7) − 2(√(21)))/7))^2 = x^2 ((36)/7) + (((3x(√7) −2(√(21)))^2 )/(49)) =x^2 7∙36 + (3x(√7) −2(√(21)) )^2 = 49x^2 14x^2 − 84x(√3) + 336=0 x_1 = 4(√3) x_2 = 2(√3)

$$\:{let}\:{AB}\:=\:{a},\:{AC}=\:{c}\:,\:{AE}\:=\:{b} \\ $$$$\:\mathrm{1}.\:{Since}\:{AD}\:{is}\:{bisector}: \\ $$$$\:\frac{{a}}{\mathrm{3}\sqrt{\mathrm{3}}}\:=\:\frac{{c}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:{a}\sqrt{\mathrm{3}}\:=\:\mathrm{3}\sqrt{\mathrm{3}}\:{c} \\ $$$$\:{a}=\mathrm{3}{c}={AB} \\ $$$$\: \\ $$$$\:\mathrm{2}.\:\bigtriangleup{AFC}\:{has}\:{a}\:{hipotenuse}\:{of}\:{c},\:{then}\: \\ $$$$\:{AF}=\frac{{c}}{\mathrm{2}}\:;\:{FC}=\:\frac{{c}\sqrt{\mathrm{3}}}{\mathrm{2}}\:\: \\ $$$$\:\mathrm{3}.\:{FB}=\:{AB}−{AF}\:\Rightarrow\:{FB}=\:\frac{\mathrm{5}{c}}{\mathrm{2}} \\ $$$$\:{FB}^{\mathrm{2}} \:+\:{FC}^{\mathrm{2}} =\:\left(\mathrm{4}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \\ $$$$\:\frac{\mathrm{25}{c}^{\mathrm{2}} }{\mathrm{4}}\:+\:\frac{\mathrm{3}{c}^{\mathrm{2}} }{\mathrm{4}}=\:\mathrm{48} \\ $$$$\:{c}=\:\sqrt{\frac{\mathrm{48}\centerdot\mathrm{4}}{\mathrm{28}}}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{3}}}{\:\sqrt{\mathrm{7}}}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{21}}}{\mathrm{7}} \\ $$$$\:\mathrm{4}.\:\bigtriangleup{AHC}\:\cong\:\bigtriangleup{AFC}\:\left({Angle};\:{side};\:{Angle}\right) \\ $$$$\:{AH}\:=\:\frac{{c}}{\mathrm{2}};\:\:{CH}=\:\frac{{c}\sqrt{\mathrm{3}}}{\mathrm{2}} \\ $$$$\:\mathrm{5}.\:\:{AC}\:{is}\:{also}\:{bisector} \\ $$$$\:\:\frac{\mathrm{3}{c}}{\mathrm{4}\sqrt{\mathrm{3}}}\:=\:\frac{{b}}{{x}} \\ $$$$\:{b}\:=\:\:\frac{\mathrm{3}{x}\sqrt{\mathrm{7}}}{\mathrm{7}}\: \\ $$$$\:\mathrm{6}.\:{HE}\:=\:{AE}\:−\:{AH}\:=\:{b}−\frac{{c}}{\mathrm{2}} \\ $$$$\:{CH}^{\mathrm{2}} \:+\:{HE}^{\mathrm{2}} ={x}^{\mathrm{2}} \\ $$$$\:\frac{\mathrm{3}{c}^{\mathrm{2}} }{\mathrm{4}}\:+\:\left({b}−\frac{{c}}{\mathrm{2}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} \\ $$$$\:\frac{\mathrm{3}}{\mathrm{4}}\centerdot\:\left(\frac{\mathrm{4}\sqrt{\mathrm{21}}}{\mathrm{7}}\right)^{\mathrm{2}} \:+\:\left(\frac{\mathrm{3}{x}\sqrt{\mathrm{7}}\:−\:\mathrm{2}\sqrt{\mathrm{21}}}{\mathrm{7}}\right)^{\mathrm{2}} =\:{x}^{\mathrm{2}} \\ $$$$\:\frac{\mathrm{36}}{\mathrm{7}}\:+\:\frac{\left(\mathrm{3}{x}\sqrt{\mathrm{7}}\:−\mathrm{2}\sqrt{\mathrm{21}}\right)^{\mathrm{2}} }{\mathrm{49}}\:={x}^{\mathrm{2}} \\ $$$$\:\mathrm{7}\centerdot\mathrm{36}\:+\:\left(\mathrm{3}{x}\sqrt{\mathrm{7}}\:−\mathrm{2}\sqrt{\mathrm{21}}\:\right)^{\mathrm{2}} =\:\mathrm{49}{x}^{\mathrm{2}} \\ $$$$\:\mathrm{14}{x}^{\mathrm{2}} \:−\:\mathrm{84}{x}\sqrt{\mathrm{3}}\:+\:\:\mathrm{336}=\mathrm{0} \\ $$$$\:{x}_{\mathrm{1}} =\:\mathrm{4}\sqrt{\mathrm{3}} \\ $$$$\:{x}_{\mathrm{2}} =\:\mathrm{2}\sqrt{\mathrm{3}} \\ $$$$\: \\ $$

Commented by HeferH last updated on 22/Oct/22

if x = 4(√3) then b= ((3∙4(√3) ∙ (√7))/7) = ((12(√(21)))/7) but AB= 3c = ((4(√(21)))/7) ∙ 3= ((12(√(21)))/7) that also implies that AC is heigth (bc its isosceles) and it doesnt satisfy: AC^2 + BC^2 = AB^2

$$\:{if}\:{x}\:=\:\mathrm{4}\sqrt{\mathrm{3}}\:{then}\:{b}=\:\frac{\mathrm{3}\centerdot\mathrm{4}\sqrt{\mathrm{3}}\:\centerdot\:\sqrt{\mathrm{7}}}{\mathrm{7}}\:=\:\frac{\mathrm{12}\sqrt{\mathrm{21}}}{\mathrm{7}}\:\: \\ $$$$\:{but}\:{AB}=\:\mathrm{3}{c}\:=\:\frac{\mathrm{4}\sqrt{\mathrm{21}}}{\mathrm{7}}\:\centerdot\:\mathrm{3}=\:\frac{\mathrm{12}\sqrt{\mathrm{21}}}{\mathrm{7}} \\ $$$$\:{that}\:{also}\:{implies}\:{that}\:{AC}\:{is}\:{heigth} \\ $$$$\:\left({bc}\:{its}\:{isosceles}\right)\:{and}\:{it}\:{doesnt}\:{satisfy}: \\ $$$$\:{AC}^{\mathrm{2}} \:+\:{BC}^{\mathrm{2}} =\:{AB}^{\mathrm{2}} \\ $$

Commented by ARUNG_Brandon_MBU last updated on 22/Oct/22

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