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Question Number 147260 by mathdanisur last updated on 19/Jul/21

Answered by Rasheed.Sindhi last updated on 19/Jul/21

$$AM=MB=x$$$$MD=y\Rightarrow DC=2y\Rightarrow MC=3y$$$$△BDM:BD=\sqrt{x^2-y^2}$$$$△BDC:BC=\sqrt{{\left(\sqrt{x^2-y^2}\right)}^2+{\left(2y\right)}^2}$$$$=\sqrt{x^2+3y^2}..............i$$$$△BMC:BC=\sqrt{{\left(3y\right)}^2-x^2}$$$$=\sqrt{9y^2-x^2}.............ii$$$$i\mathrm{\&}ii:\sqrt{x^2+3y^2}=\sqrt{9y^2-x^2}$$$$x^2+3y^2=9y^2-x^2$$$$2x^2=6y^2\Rightarrow x=\sqrt{3}y$$$$$$$$△ABC:AC=\sqrt{{\left(2x\right)}^2+{\left(\sqrt{x^2+3y^2}\right)}^2}$$$$=\sqrt{5x^2+3y^2}=\sqrt{5{\left(\sqrt{3}y\right)}^2+3y^2}$$$$=3\sqrt{2}y$$$$i:BC=\sqrt{x^2+3y^2}=\sqrt{{\left(\sqrt{3}y\right)}^2+3y^2}$$$$=\sqrt{6}y$$$$AB:AC=2x:3\sqrt{2}y$$$$=2\left(\sqrt{3}y\right):3\sqrt{2}y=\frac{2\sqrt{3}}{3\sqrt{2}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{2\sqrt{6}}{6}$$

Commented by mathdanisur last updated on 20/Jul/21

$$thankyouSer$$

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