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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

$$2\sqrt{3}?$$

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

$$Thentheansweris2\sqrt{3},Ididitwiththe$$$$bisectortheoremandpythagoras,{you}^{\prime }ll$$$$getacuadraticequation,wediscardone$$$$ofthesolutions\left(4\sqrt{3}\right)becauseit{wouldn}^{\prime }t$$$$satisfytheotherequation.$$$$$$

Commented by HeferH last updated on 22/Oct/22

$$ItsADbisector?$$

Commented by ARUNG_Brandon_MBU last updated on 22/Oct/22

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

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

$$letAB=a,AC=c,AE=b$$$$1.SinceADisbisector:$$$$\frac{a}{3\sqrt{3}}=\frac{c}{\sqrt{3}}$$$$a\sqrt{3}=3\sqrt{3}c$$$$a=3c=AB$$$$$$$$2.△AFChasahipotenuseofc,then$$$$AF=\frac{c}{2};FC=\frac{c\sqrt{3}}{2}$$$$3.FB=AB-AF\Rightarrow FB=\frac{5c}{2}$$$${FB}^2+{FC}^2={\left(4\sqrt{3}\right)}^2$$$$\frac{25c^2}{4}+\frac{3c^2}{4}=48$$$$c=\sqrt{\frac{48\cdot 4}{28}}=\frac{4\sqrt{3}}{\sqrt{7}}=\frac{4\sqrt{21}}{7}$$$$4.△AHC\cong △AFC(Angle;side;Angle)$$$$AH=\frac{c}{2};CH=\frac{c\sqrt{3}}{2}$$$$5.ACisalsobisector$$$$\frac{3c}{4\sqrt{3}}=\frac{b}{x}$$$$b=\frac{3x\sqrt{7}}{7}$$$$6.HE=AE-AH=b-\frac{c}{2}$$$${CH}^2+{HE}^2=x^2$$$$\frac{3c^2}{4}+{(b-\frac{c}{2})}^2=x^2$$$$\frac{3}{4}\cdot {\left(\frac{4\sqrt{21}}{7}\right)}^2+{\left(\frac{3x\sqrt{7}-2\sqrt{21}}{7}\right)}^2=x^2$$$$\frac{36}{7}+\frac{{(3x\sqrt{7}-2\sqrt{21})}^2}{49}=x^2$$$$7\cdot 36+{(3x\sqrt{7}-2\sqrt{21})}^2=49x^2$$$$14x^2-84x\sqrt{3}+336=0$$$$x_1=4\sqrt{3}$$$$x_2=2\sqrt{3}$$$$$$

Commented by HeferH last updated on 22/Oct/22

$$ifx=4\sqrt{3}thenb=\frac{3\cdot 4\sqrt{3}\cdot \sqrt{7}}{7}=\frac{12\sqrt{21}}{7}$$$$butAB=3c=\frac{4\sqrt{21}}{7}\cdot 3=\frac{12\sqrt{21}}{7}$$$$thatalsoimpliesthatACisheigth$$$$\left(bcitsisosceles\right)anditdoesntsatisfy:$$$${AC}^2+{BC}^2={AB}^2$$

Commented by ARUNG_Brandon_MBU last updated on 22/Oct/22

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