Question Number 22518 by ajfour last updated on 19/Oct/17
Commented by ajfour last updated on 19/Oct/17
$${Q}.\mathrm{22515}\:\left({solution}\right) \\ $$
Answered by ajfour last updated on 19/Oct/17
$$\left(\frac{{AD}.{BC}}{{AB}.{CD}}\right)^{\mathrm{2}} =\frac{\left({a}^{\mathrm{2}} +{c}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta\right)\left({a}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta+{c}^{\mathrm{2}} \right)}{\left({a}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{2}} \theta\right)\left({c}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{2}} \theta\right)} \\ $$$$=\left(\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\mathrm{cos}\:^{\mathrm{2}} \theta+\mathrm{sin}\:^{\mathrm{2}} \theta\right)\left(\mathrm{sin}\:^{\mathrm{2}} \theta+\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\mathrm{cos}\:^{\mathrm{2}} \theta\right) \\ $$$$=\left(\mathrm{sin}\:^{\mathrm{4}} \theta+\mathrm{cos}\:^{\mathrm{4}} \theta+\mathrm{2sin}\:^{\mathrm{2}} \theta\:\mathrm{cos}\:^{\mathrm{2}} \theta\right) \\ $$$$\:\:\:\:\:\:\:+\left(\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mathrm{2}+\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }\right)\mathrm{sin}\:^{\mathrm{2}} \theta\:\mathrm{cos}\:^{\mathrm{2}} \theta \\ $$$$=\mathrm{1}+\left(\frac{{c}}{{a}}−\frac{{a}}{{c}}\right)^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta\:\mathrm{cos}\:^{\mathrm{2}} \theta\:\geqslant\:\mathrm{1}\: \\ $$$${Hence} \\ $$$$\:{AD}.{BC}\:\geqslant\:{AB}.{CD} \\ $$$$ \\ $$
Commented by Tinkutara last updated on 20/Oct/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Answered by ajfour last updated on 19/Oct/17
![((AD+BC)/(AB+CD)) =(((√(a^2 +c^2 tan^2 θ)) +(√(c^2 +a^2 tan^2 θ)))/((a+c)sec θ)) =(1/((a+c))) [(√(a^2 cos^2 θ+c^2 sin^2 θ)) + (√(c^2 cos^2 θ+a^2 sin^2 θ)) ] ......](https://www.tinkutara.com/question/Q22521.png)
$$\frac{{AD}+{BC}}{{AB}+{CD}}\:=\frac{\sqrt{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta}\:+\sqrt{{c}^{\mathrm{2}} +{a}^{\mathrm{2}} \mathrm{tan}\:^{\mathrm{2}} \theta}}{\left({a}+{c}\right)\mathrm{sec}\:\theta} \\ $$$$=\frac{\mathrm{1}}{\left({a}+{c}\right)}\:\left[\sqrt{{a}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta+{c}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}\:+\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\sqrt{{c}^{\mathrm{2}} \mathrm{cos}\:^{\mathrm{2}} \theta+{a}^{\mathrm{2}} \mathrm{sin}\:^{\mathrm{2}} \theta}\:\right] \\ $$$$…… \\ $$
Commented by mrW1 last updated on 19/Oct/17
Commented by mrW1 last updated on 19/Oct/17
$$\mathrm{AA}'//\mathrm{CD}//\mathrm{AB} \\ $$$$\mathrm{AA}''//\mathrm{CB} \\ $$$$\mathrm{DA}'//\mathrm{AC}//\mathrm{BA}'' \\ $$$$\mathrm{DA}'=\mathrm{AC}=\mathrm{BA}'' \\ $$$$\mathrm{AB}+\mathrm{CD}=\mathrm{AB}+\mathrm{AA}'=\mathrm{A}'\mathrm{B}=\mathrm{DA}''\leqslant\mathrm{AD}+\mathrm{AA}''=\mathrm{AD}+\mathrm{BC} \\ $$
Commented by Tinkutara last updated on 20/Oct/17
$$\mathrm{Why}\:\mathrm{in}\:\mathrm{last}\:\mathrm{line}\:\mathrm{A}'\mathrm{B}=\mathrm{DA}''? \\ $$
Commented by mrW1 last updated on 20/Oct/17
$$\mathrm{Since}\:\mathrm{DA}'//\mathrm{CA}//\mathrm{BA}''\:\mathrm{and}\:\mathrm{AC}\bot\mathrm{DB}, \\ $$$$\Rightarrow\Delta\mathrm{A}'\mathrm{DB}\:\mathrm{and}\:\Delta\mathrm{A}''\mathrm{BD}\:\mathrm{are}\:\mathrm{right}\:\mathrm{angled}. \\ $$$$\mathrm{Since}\:\mathrm{DA}'=\mathrm{BA}''=\mathrm{CA}\:\mathrm{and}\:\mathrm{DB}=\mathrm{BD}, \\ $$$$\Rightarrow\mathrm{BA}'=\mathrm{DA}'' \\ $$
Commented by Tinkutara last updated on 20/Oct/17
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$
Commented by ajfour last updated on 20/Oct/17
$${wonderful}\:{sir},\:{thoughtful} \\ $$$${constructions}. \\ $$