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Question Number 135128 by bramlexs22 last updated on 10/Mar/21

Commented by mr W last updated on 10/Mar/21

$$isABCDasquare?$$

Commented by mr W last updated on 10/Mar/21

Commented by bramlexs22 last updated on 10/Mar/21

$$\mathrm{yes}$$

Commented by bramlexs22 last updated on 10/Mar/21

$$\mathrm{is}\measuredangle \mathrm{AEB}=135°$$

Answered by mr W last updated on 10/Mar/21

$$a=sidelength$$$$\cos \mathrm{\angle }ABE=\frac{2^2+a^2-1^2}{2\times 2\times a}=\frac{a^2+3}{4a}$$$$\cos \mathrm{\angle }CBE=\frac{2^2+a^2-3^2}{2\times 2\times a}=\frac{a^2-5}{4a}=\sin \mathrm{\angle }ABE$$$$\Rightarrow {\left(\frac{a^2+3}{4a}\right)}^2+{\left(\frac{a^2-5}{4a}\right)}^2=1$$$$\Rightarrow a^4+6a^2+9+a^4-10a^2+25=16a^2$$$$\Rightarrow a^4-10a^2+17=0$$$$\Rightarrow a^2=5\pm 2\sqrt{2}$$$$\Rightarrow a=\sqrt{5\pm 2\sqrt{2}}$$$$\cos \mathrm{\angle }AEB=\frac{1^2+2^2-a^2}{2\times 1\times 2}=\frac{5-5\pm 2\sqrt{2}}{4}=\pm \frac{\sqrt{2}}{2}$$$$\Rightarrow \mathrm{\angle }AEB=45°or135°$$

Commented by EDWIN88 last updated on 10/Mar/21

I prefer the angle is obtuse sir

Commented by bramlexs22 last updated on 10/Mar/21

$$\mathrm{yes}...$$

Commented by mr W last updated on 10/Mar/21

$$itisnotrequestedthatEshouldlie$$$$insidethesquare,sotherearetwo$$$$possibilities.$$

Commented by mr W last updated on 10/Mar/21

Commented by mr W last updated on 10/Mar/21

$$thecircleshaveradiifrom1,2,3.$$$$inbothcaseswehave$$$$AE:BE:CE=1:2:3$$

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