Question Number 39508 by behi83417@gmail.com last updated on 06/Jul/18
Commented by behi83417@gmail.com last updated on 06/Jul/18
$${R}\boldsymbol{{ight}}\:\boldsymbol{{angele}}:\boldsymbol{{A}}\overset{\bigtriangleup} {\boldsymbol{{B}A}}',\left(\measuredangle\boldsymbol{{B}}=\mathrm{90}^{\bullet} \right)\:\boldsymbol{{is}} \\ $$$$\boldsymbol{{given}}. \\ $$$$\boldsymbol{{red}}\:\boldsymbol{{lines}}\:\boldsymbol{{are}}\:\boldsymbol{{angular}}\:\boldsymbol{{bisectors}}\:\boldsymbol{{of}} \\ $$$$\measuredangle\boldsymbol{{A}},\measuredangle\boldsymbol{{A}}'. \\ $$$$\left.\mathrm{1}\right)\boldsymbol{{show}}\:\boldsymbol{{that}}:\boldsymbol{{S}}_{\boldsymbol{{A}}\overset{\lozenge} {\boldsymbol{{D}CA}}'} =\mathrm{2}\boldsymbol{{S}}_{\boldsymbol{{A}}\overset{\bigtriangleup} {\boldsymbol{{E}A}}'} \\ $$$$\left.\mathrm{2}\right)\frac{\boldsymbol{{AA}}'}{\boldsymbol{{DC}}}=?,\frac{\boldsymbol{{S}}_{\boldsymbol{{AEA}}'} }{\boldsymbol{{S}}_{\boldsymbol{{D}}\overset{\bigtriangleup} {\boldsymbol{{E}C}}} }=?. \\ $$
Answered by MrW3 last updated on 07/Jul/18
Commented by MrW3 last updated on 07/Jul/18
$$\angle{A}'+\angle{A}=\mathrm{90}° \\ $$$${A}'{C}'={A}'{C} \\ $$$${AD}'={AD} \\ $$$$\Rightarrow{C}'{E}={CE},\:{D}'{E}={DE} \\ $$$$\angle{CED}=\angle{A}'{EA}=\mathrm{180}°−\frac{\angle{A}'}{\mathrm{2}}−\frac{\angle{A}}{\mathrm{2}}=\mathrm{135}° \\ $$$$\alpha=\mathrm{180}°−\mathrm{135}°=\mathrm{45}° \\ $$$$\angle{C}'{ED}'=\mathrm{135}°−\mathrm{2}\alpha=\mathrm{45}° \\ $$$${A}_{\Delta{CED}} =\frac{\mathrm{1}}{\mathrm{2}}×{CE}×{DE}×\mathrm{sin}\:\angle{CED} \\ $$$${A}_{\Delta{C}'{ED}'} =\frac{\mathrm{1}}{\mathrm{2}}×{C}'{E}×{D}'{E}×\mathrm{sin}\:\angle{C}'{ED}' \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{CE}×{DE}×\mathrm{sin}\:\mathrm{45}° \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{CE}×{DE}×\mathrm{sin}\:\mathrm{135}° \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}×{CE}×{DE}×\mathrm{sin}\:\angle{CED} \\ $$$$={A}_{\Delta{CED}} \\ $$$${A}_{\Delta{A}'{EA}} ={A}_{\Delta{A}'{EC}'} +{A}_{\Delta{C}'{ED}'} +{A}_{\Delta{D}'{EA}} \\ $$$$={A}_{\Delta{A}'{EC}} +{A}_{\Delta{CED}} +{A}_{\Delta{DEA}} \\ $$$${A}_{{A}'{CDA}} ={A}_{\Delta{A}'{EA}} +{A}_{\Delta{A}'{EC}} +{A}_{\Delta{CED}} +{A}_{\Delta{DEA}} \\ $$$$=\mathrm{2}{A}_{\Delta{A}'{EA}} \\ $$
Commented by behi83417@gmail.com last updated on 07/Jul/18
$${sweet}\:{waiting}\:{and}\:{nice}\:{proof}.{thanks} \\ $$$${my}\:{master}. \\ $$
Commented by behi83417@gmail.com last updated on 07/Jul/18
$${perpendicular}\:{from}\::{E}\:{to}\:{AA}',{say}:{EH}. \\ $$$${S}_{{AEA}'} =\frac{\mathrm{1}}{\mathrm{2}}.{EH}.{AA}'=\frac{\mathrm{1}}{\mathrm{2}}.{r}.{b} \\ $$$$\left[\boldsymbol{{r}}=\boldsymbol{{radius}}\:\boldsymbol{{of}}\:\boldsymbol{{incircle}}\:\boldsymbol{{of}}\:\boldsymbol{{A}}\overset{\bigtriangleup} {\boldsymbol{{B}A}}'.\right] \\ $$$$\frac{{sin}\frac{{A}}{\mathrm{2}}}{{A}'{C}}=\frac{{sinACA}'}{{b}},\frac{{sin}\frac{{A}}{\mathrm{2}}}{{BC}}=\frac{{sinACB}}{{c}} \\ $$$$\frac{{b}}{{A}'{C}}=\frac{{c}}{{BC}}\Rightarrow\frac{{A}'{C}}{{BC}}=\frac{{b}}{{c}}\Rightarrow\frac{{A}'{C}}{{a}}=\frac{{b}}{{b}+{c}} \\ $$$$\Rightarrow{A}'{C}=\frac{{ab}}{{b}+{c}},{BC}=\frac{{ac}}{{b}+{c}},{BD}=\frac{{ca}}{{b}+{a}},{AD}=\frac{{cb}}{{b}+{a}} \\ $$$${S}_{{ADCA}'} ={S}_{{ABA}'} −{S}_{{DBC}} =\frac{\mathrm{1}}{\mathrm{2}}{ac}−\frac{\mathrm{1}}{\mathrm{2}}{BD}.{BC}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ac}−\frac{\mathrm{1}}{\mathrm{2}}.\frac{{ac}}{{a}+{b}}.\frac{{ac}}{{b}+{c}}=\frac{{ac}}{\mathrm{2}}\left(\mathrm{1}−\frac{{ac}}{\left({a}+{b}\right)\left({b}+{c}\right)}\right)= \\ $$$$=\frac{{ac}}{\mathrm{2}}\left(\frac{{ab}+{ac}+{b}^{\mathrm{2}} +{bc}−{ac}}{\left({a}+{b}\right)\left({b}+{c}\right)}\right)=\frac{{ac}}{\mathrm{2}}.\frac{{b}^{\mathrm{2}} +{ab}+{bc}}{\left({a}+{b}\right)\left({b}+{c}\right)}= \\ $$$$=\frac{{ac}}{\mathrm{2}}.\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +{b}^{\mathrm{2}} +\mathrm{2}{ba}+\mathrm{2}{bc}}{\mathrm{2}\left({a}+{b}\right)\left({b}+{c}\right)}=\frac{{ac}}{\mathrm{2}}.\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}{ac}}{\mathrm{2}{ab}+\mathrm{2}{ac}+\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{bc}}= \\ $$$$={S}.\frac{\left(\mathrm{2}{p}\right)^{\mathrm{2}} −\mathrm{4}{S}}{\left(\mathrm{2}{p}\right)^{\mathrm{2}} }={S}.\frac{{p}^{\mathrm{2}} −{S}}{{p}^{\mathrm{2}} }={S}−\frac{{S}^{\mathrm{2}} }{{p}^{\mathrm{2}} }={S}−{r}^{\mathrm{2}} = \\ $$$$={r}.{p}−{r}^{\mathrm{2}} ={r}.\left({p}−{r}\right)={r}.{b}=\mathrm{2}{S}_{{AEA}'} .\:\:\blacksquare \\ $$$${note}:{in}\:{right}\:{angle}\:{triangle}:\boldsymbol{{b}}=\boldsymbol{{p}}−\boldsymbol{{r}} \\ $$$$\:{because}:\:\:{p}−{r}={p}−\frac{{S}}{{p}}=\frac{{p}^{\mathrm{2}} −{S}}{{p}}= \\ $$$$=\frac{\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}{ac}}{\mathrm{4}{p}}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ab}+\mathrm{2}{bc}}{\mathrm{4}{p}} \\ $$$$=\frac{\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{ba}+\mathrm{2}{bc}}{\mathrm{4}{p}}=\frac{\mathrm{2}{b}\left({a}+{b}+{c}\right)}{\mathrm{4}{p}}=\frac{\mathrm{2}{b}.\mathrm{2}{p}}{\mathrm{4}{p}}=\boldsymbol{{b}}. \\ $$
Commented by behi83417@gmail.com last updated on 07/Jul/18
$${DC}^{\mathrm{2}} ={DB}^{\mathrm{2}} +{CB}^{\mathrm{2}} =\left(\frac{{ac}}{{a}+{b}}\right)^{\mathrm{2}} +\left(\frac{{ac}}{{c}+{b}}\right)^{\mathrm{2}} = \\ $$$${a}^{\mathrm{2}} {c}^{\mathrm{2}} .\frac{\left({b}+{c}\right)^{\mathrm{2}} +\left({b}+{a}\right)^{\mathrm{2}} }{\left({a}+{b}\right)^{\mathrm{2}} \left({c}+{b}\right)^{\mathrm{2}} }=\frac{\mathrm{4}{S}^{\mathrm{2}} {b}\left({b}+{p}\right)}{\mathrm{4}{p}^{\mathrm{4}} }= \\ $$$$=\frac{{S}^{\mathrm{2}} }{{p}^{\mathrm{2}} }.{b}.\frac{{b}+{p}}{{p}^{\mathrm{2}} }={b}.{r}^{\mathrm{2}} .\frac{\mathrm{2}{p}−{r}}{{p}^{\mathrm{2}} }={b}.\frac{{r}^{\mathrm{2}} }{{p}^{\mathrm{2}} }\left(\mathrm{2}{p}−{r}\right) \\ $$$$\Rightarrow\frac{{DC}}{{AA}'}=\frac{{r}^{\mathrm{2}} }{{p}^{\mathrm{2}} }\left(\mathrm{2}{p}−{r}\right). \\ $$$${note}: \\ $$$$\left({b}+{c}\right)^{\mathrm{2}} +\left({b}+{a}\right)^{\mathrm{2}} =\mathrm{3}{b}^{\mathrm{2}} +\mathrm{2}{bc}+\mathrm{2}{ab}= \\ $$$${b}\left(\mathrm{3}{b}+{a}+{c}\right)={b}\left(\mathrm{2}{b}+\mathrm{2}{p}\right)=\mathrm{2}{b}\left({b}+{p}\right) \\ $$$$\left({a}+{b}\right)\left({c}+{b}\right)={ac}+{ab}+{bc}+{b}^{\mathrm{2}} = \\ $$$$=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}+\mathrm{2}{bc}+\mathrm{2}{ab}}{\mathrm{2}}=\frac{\left(\mathrm{2}{p}\right)^{\mathrm{2}} }{\mathrm{2}}=\mathrm{2}{p}^{\mathrm{2}} \\ $$
Commented by behi83417@gmail.com last updated on 08/Jul/18
$${S}_{{ACE}} =\frac{\mathrm{1}}{\mathrm{2}}.{r}.{A}'{C}=\frac{\mathrm{1}}{\mathrm{2}}.{r}.\frac{{ab}}{{b}+{c}} \\ $$$${S}_{{ADE}} =\frac{\mathrm{1}}{\mathrm{2}}.{r}.{AD}=\frac{\mathrm{1}}{\mathrm{2}}.{r}.\frac{{c}.{b}}{{a}+{b}} \\ $$$${S}_{{DEC}} =\mathrm{2}{S}_{{AEA}'} −{S}_{{AEA}'} −\frac{{b}.{r}}{\mathrm{2}}\left(\frac{{a}}{{b}+{c}}+\frac{{c}}{{a}+{b}}\right)= \\ $$$$=\frac{{b}.{r}}{\mathrm{2}}\left(\mathrm{1}−\frac{{a}}{{b}+{c}}−\frac{{c}}{{a}+{b}}\right)=\frac{{br}\left({ab}+{ac}+{bc}+{b}^{\mathrm{2}} −{a}^{\mathrm{2}} −{ab}−{c}^{\mathrm{2}} −{cb}\right)}{\mathrm{2}\left({b}+{c}\right)\left({b}+{a}\right)}= \\ $$$$=\frac{{br}.{ac}}{\mathrm{4}{p}^{\mathrm{2}} }=\frac{{br}.\mathrm{2}{S}}{\mathrm{4}{p}^{\mathrm{2}} }=\frac{{b}}{\mathrm{2}{p}} \\ $$$$\Rightarrow\frac{{S}_{{AEA}'} }{{S}_{{DRC}} }=\frac{\frac{{br}}{\mathrm{2}}}{\frac{{b}}{\mathrm{2}{p}}}={r}.{p}\:\:\:\left(????????????\rightarrow\infty\right) \\ $$