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Question Number 22518 by ajfour last updated on 19/Oct/17

Commented by ajfour last updated on 19/Oct/17

$$Q.22515\left(solution\right)$$

Answered by ajfour last updated on 19/Oct/17

$${\left(\frac{AD.BC}{AB.CD}\right)}^2=\frac{(a^2+c^2\tan {}^2\theta )(a^2\tan {}^2\theta +c^2)}{\left(a^2\sec {}^2\theta \right)\left(c^2\sec {}^2\theta \right)}$$$$=(\frac{a^2}{c^2}\cos {}^2\theta +\sin {}^2\theta )(\sin {}^2\theta +\frac{c^2}{a^2}\cos {}^2\theta )$$$$=(\sin {}^4\theta +\cos {}^4\theta +2\sin {}^2\theta \cos {}^2\theta )$$$$+(\frac{c^2}{a^2}-2+\frac{a^2}{c^2})\sin {}^2\theta \cos {}^2\theta$$$$=1+{(\frac{c}{a}-\frac{a}{c})}^2\sin {}^2\theta \cos {}^2\theta ⩾1$$$$Hence$$$$AD.BC⩾AB.CD$$$$$$

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

$$\frac{AD+BC}{AB+CD}=\frac{\sqrt{a^2+c^2\tan {}^2\theta }+\sqrt{c^2+a^2\tan {}^2\theta }}{(a+c)\sec \theta }$$$$=\frac{1}{(a+c)}[\sqrt{a^2\cos {}^2\theta +c^2\sin {}^2\theta }+$$$$\sqrt{c^2\cos {}^2\theta +a^2\sin {}^2\theta }]$$$$......$$

Commented by mrW1 last updated on 19/Oct/17

Commented by mrW1 last updated on 19/Oct/17

$${\mathrm{AA}}^{\prime }//\mathrm{CD}//\mathrm{AB}$$$${\mathrm{AA}}^{″}//\mathrm{CB}$$$${\mathrm{DA}}^{\prime }//\mathrm{AC}//{\mathrm{BA}}^{″}$$$${\mathrm{DA}}^{\prime }=\mathrm{AC}={\mathrm{BA}}^{″}$$$$\mathrm{AB}+\mathrm{CD}=\mathrm{AB}+{\mathrm{AA}}^{\prime }={\mathrm{A}}^{\prime }\mathrm{B}={\mathrm{DA}}^{″}⩽\mathrm{AD}+{\mathrm{AA}}^{″}=\mathrm{AD}+\mathrm{BC}$$

Commented by Tinkutara last updated on 20/Oct/17

$$\mathrm{Why}\mathrm{in}\mathrm{last}\mathrm{line}{\mathrm{A}}^{\prime }\mathrm{B}={\mathrm{DA}}^{″}?$$

Commented by mrW1 last updated on 20/Oct/17

$$\mathrm{Since}{\mathrm{DA}}^{\prime }//\mathrm{CA}//{\mathrm{BA}}^{″}\mathrm{and}\mathrm{AC}\mathrm{\perp }\mathrm{DB},$$$$\Rightarrow \mathrm{\Delta }{\mathrm{A}}^{\prime }\mathrm{DB}\mathrm{and}\mathrm{\Delta }{\mathrm{A}}^{″}\mathrm{BD}\mathrm{are}\mathrm{right}\mathrm{angled}.$$$$\mathrm{Since}{\mathrm{DA}}^{\prime }={\mathrm{BA}}^{″}=\mathrm{CA}\mathrm{and}\mathrm{DB}=\mathrm{BD},$$$$\Rightarrow {\mathrm{BA}}^{\prime }={\mathrm{DA}}^{″}$$

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

$$wonderfulsir,thoughtful$$$$constructions.$$

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