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Question Number 12127 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 13/Apr/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17

$$intriangle:ABCwehave:$$$$AB=m,AC=n,AD=\frac{\sqrt{2}}{2},$$$$BD\mathrm{\perp }DC,\measuredangle BDA=\measuredangle ADC.$$$$find:BDandDCintermof:m,n.$$

Answered by mrW1 last updated on 14/Apr/17

$$DC=x$$$$DB=y$$$$\mathrm{\angle }ADC=\mathrm{\angle }ADB=\frac{1}{2}(360-90)=135°$$$${AC}^2={AD}^2+{DC}^2-2\times AD\times DC\times \cos \mathrm{\angle }ADC$$$$n^2={\left(\frac{\sqrt{2}}{2}\right)}^2+x^2-2\times \frac{\sqrt{2}}{2}\times x\times \cos 135°$$$$n^2={\left(\frac{\sqrt{2}}{2}\right)}^2+x^2+2\times \frac{\sqrt{2}}{2}\times x\times \frac{\sqrt{2}}{2}$$$$n^2=\frac{1}{2}+x^2+x$$$$x^2+x+\frac{1}{2}-n^2=0$$$$x=DC=\frac{-1+\sqrt{1-4(\frac{1}{2}-n^2)}}{2}=\frac{\sqrt{4n^2-1}-1}{2}$$$$similarily$$$$y=DB=\frac{\sqrt{4m^2-1}-1}{2}$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17

$$thankyouverymuch,dearmrW1.$$

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