Question-222662

Question Number 222662 by MrGaster last updated on 04/Jul/25

Commented by MrGaster last updated on 04/Jul/25

∃△ABC,∠B=90°,BA=AC,BD=3∧∠CED=45°,CE=(√2)AE,CE=?

$$\exists\bigtriangleup{ABC},\angle{B}=\mathrm{90}°,{BA}={AC},{BD}=\mathrm{3}\wedge\angle{CED}=\mathrm{45}°,{CE}=\sqrt{\mathrm{2}}{AE},{CE}=? \\ $$

Commented by mr W last updated on 04/Jul/25

$${BA}={AC}\:\:??? \\ $$

Commented by gabthemathguy25 last updated on 04/Jul/25

$${uh}\:{is}\:{this}\:{a}\:\mathrm{45}°,\mathrm{45}°,\mathrm{90}°\:{triangle}? \\ $$

Commented by MrGaster last updated on 04/Jul/25

Extend ae, cross C and make cf perpendicular to ae. In the right triangle cef, e is 45 degrees, so ce= root number 2 times ef, so ae=ef=cf=x, ac is root number 5 times x, other sides can be represented by x, and then Pythagorean theorem can solve X.

Answered by gabthemathguy25 last updated on 06/Jul/25

AB=BC AC=(√(x^2 +x^2 ))=(√2)x let B=(0,0) A = (0,x) C=(x,0) lets assume D=(3,0) x=BC=3+DC im solving this later ∠CED = 45°, CE =(√2)∙AE using the law of cosines let AE=a, so CE = (√2)a ⇒cos(∠CEA)=((a^2 +2a^2 −2x^2 )/(2a∙(√2)a))=((3a^2 −2x^2 )/(2a^2 (√2))) thats complicated, im going to try something different im going to solve this manually E=(2,4) AE=(√((2)^2 +(4−6)^2 ))=(√(4+4))=(√8) CE=(√((2−6)^2 +4^2 ))=(√(16+16))=(√(32)) ⇒CE=(√(32))=(√2)∙(√(16))=(√2)∙4 AE=(√8)=2(√2) CE=(√(32))=4(√2)

$${AB}={BC} \\ $$$${AC}=\sqrt{{x}^{\mathrm{2}} +{x}^{\mathrm{2}} }=\sqrt{\mathrm{2}}{x} \\ $$$$\mathrm{let}\:{B}=\left(\mathrm{0},\mathrm{0}\right)\:{A}\:=\:\left(\mathrm{0},{x}\right)\:{C}=\left({x},\mathrm{0}\right)\: \\ $$$$\mathrm{lets}\:\mathrm{assume}\:{D}=\left(\mathrm{3},\mathrm{0}\right)\: \\ $$$${x}={BC}=\mathrm{3}+{DC}\:{im}\:{solving}\:{this}\:{later} \\ $$$$\angle{CED}\:=\:\mathrm{45}°,\:{CE}\:=\sqrt{\mathrm{2}}\centerdot{AE} \\ $$$$\mathrm{using}\:\mathrm{the}\:\mathrm{law}\:\mathrm{of}\:\mathrm{cosines}\:\mathrm{let}\:{AE}={a},\:{so}\:{CE}\:=\:\sqrt{\mathrm{2}}{a} \\ $$$$\Rightarrow\mathrm{cos}\left(\angle{CEA}\right)=\frac{{a}^{\mathrm{2}} +\mathrm{2}{a}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}{a}\centerdot\sqrt{\mathrm{2}}{a}}=\frac{\mathrm{3}{a}^{\mathrm{2}} −\mathrm{2}{x}^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} \sqrt{\mathrm{2}}} \\ $$$${thats}\:{complicated},\:{im}\:{going}\:{to}\:{try}\:{something}\:{different} \\ $$$${im}\:{going}\:{to}\:{solve}\:{this}\:{manually} \\ $$$${E}=\left(\mathrm{2},\mathrm{4}\right) \\ $$$${AE}=\sqrt{\left(\mathrm{2}\right)^{\mathrm{2}} +\left(\mathrm{4}−\mathrm{6}\right)^{\mathrm{2}} }=\sqrt{\mathrm{4}+\mathrm{4}}=\sqrt{\mathrm{8}} \\ $$$${CE}=\sqrt{\left(\mathrm{2}−\mathrm{6}\right)^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} }=\sqrt{\mathrm{16}+\mathrm{16}}=\sqrt{\mathrm{32}} \\ $$$$\Rightarrow{CE}=\sqrt{\mathrm{32}}=\sqrt{\mathrm{2}}\centerdot\sqrt{\mathrm{16}}=\sqrt{\mathrm{2}}\centerdot\mathrm{4} \\ $$$${AE}=\sqrt{\mathrm{8}}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${CE}=\sqrt{\mathrm{32}}=\mathrm{4}\sqrt{\mathrm{2}} \\ $$

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