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Question-12127




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
in triangle:ABC we have:  AB=m,AC=n,AD=((√2)/2),  BD⊥DC,∡BDA=∡ADC.  find:BD and DC in term of: m,n.
$${in}\:{triangle}:{ABC}\:{we}\:{have}: \\ $$$${AB}={m},{AC}={n},{AD}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}, \\ $$$${BD}\bot{DC},\measuredangle{BDA}=\measuredangle{ADC}. \\ $$$${find}:{BD}\:{and}\:{DC}\:{in}\:{term}\:{of}:\:{m},{n}. \\ $$
Answered by mrW1 last updated on 14/Apr/17
DC=x  DB=y  ∠ADC=∠ADB=(1/2)(360−90)=135°  AC^2 =AD^2 +DC^2 −2×AD×DC×cos ∠ADC  n^2 =(((√2)/2))^2 +x^2 −2×((√2)/2)×x×cos 135°  n^2 =(((√2)/2))^2 +x^2 +2×((√2)/2)×x×((√2)/2)  n^2 =(1/2)+x^2 +x  x^2 +x+(1/2)−n^2 =0  x=DC=((−1+(√(1−4((1/2)−n^2 ))))/2)=(((√(4n^2 −1))−1)/2)  similarily  y=DB=(((√(4m^2 −1))−1)/2)
$${DC}={x} \\ $$$${DB}={y} \\ $$$$\angle{ADC}=\angle{ADB}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{360}−\mathrm{90}\right)=\mathrm{135}° \\ $$$${AC}^{\mathrm{2}} ={AD}^{\mathrm{2}} +{DC}^{\mathrm{2}} −\mathrm{2}×{AD}×{DC}×\mathrm{cos}\:\angle{ADC} \\ $$$${n}^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}×{x}×\mathrm{cos}\:\mathrm{135}° \\ $$$${n}^{\mathrm{2}} =\left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\right)^{\mathrm{2}} +{x}^{\mathrm{2}} +\mathrm{2}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}×{x}×\frac{\sqrt{\mathrm{2}}}{\mathrm{2}} \\ $$$${n}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{2}}+{x}^{\mathrm{2}} +{x} \\ $$$${x}^{\mathrm{2}} +{x}+\frac{\mathrm{1}}{\mathrm{2}}−{n}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}={DC}=\frac{−\mathrm{1}+\sqrt{\mathrm{1}−\mathrm{4}\left(\frac{\mathrm{1}}{\mathrm{2}}−{n}^{\mathrm{2}} \right)}}{\mathrm{2}}=\frac{\sqrt{\mathrm{4}{n}^{\mathrm{2}} −\mathrm{1}}−\mathrm{1}}{\mathrm{2}} \\ $$$${similarily} \\ $$$${y}={DB}=\frac{\sqrt{\mathrm{4}{m}^{\mathrm{2}} −\mathrm{1}}−\mathrm{1}}{\mathrm{2}} \\ $$
Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 14/Apr/17
thank you very much, dear mrW1.
$${thank}\:{you}\:{very}\:{much},\:{dear}\:{mrW}\mathrm{1}. \\ $$

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