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Question Number 16409 by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Jun/17

$$\mathrm{\Delta }ABC,isequilateraland{}^{\prime }{D}^{\prime },isapointoncircumcircleofA\stackrel{\mathrm{\Delta }}{BC}.$$$$provethat:$$$$DA=DB+DC.$$

Commented by mrW1 last updated on 21/Jun/17

$$...\mathrm{and}\mathrm{\Delta }\mathrm{ABC}\mathrm{is}\mathrm{equilateral}.$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 21/Jun/17

$$yes!youarerightdearmaster.$$

Commented by RasheedSoomro last updated on 22/Jun/17

$$\mathrm{D}\mathrm{is}\mathbf{any}\mathrm{point}\mathrm{of}\mathrm{cercumference}$$$$\mathrm{If}\mathrm{D}\mathrm{is}\mathrm{beyond}\stackrel{⌢}{\mathrm{CB}}\mathrm{say}\mathrm{it}\mathrm{is}\mathrm{on}\stackrel{⌢}{\mathrm{AB}}$$$$\mathrm{then}\mathrm{how}$$$$\mathrm{DA}=\mathrm{DB}+\mathrm{DC}?$$$$\mathrm{How}?\mathrm{pl}\mathrm{explain}.$$$$$$

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17

Commented by RasheedSoomro last updated on 22/Jun/17

$$\mathrm{This}\mathrm{means}:\mathrm{For}\mathrm{the}\mathrm{result}\mathrm{DA}=\mathrm{DB}+\mathrm{DC}$$$$\mathrm{D}\mathrm{must}\mathrm{be}\mathrm{on}\stackrel{⌢}{\mathrm{BC}}\mathrm{and}$$$$\mathrm{for}\mathrm{the}\mathrm{result}\mathrm{DC}=\mathrm{DA}+\mathrm{DB}\mathrm{D}\mathrm{must}\mathrm{be}\mathrm{on}$$$$\stackrel{⌢}{\mathrm{AB}}.$$$$\mathrm{If}\mathrm{we}\mathrm{take}\mathrm{the}\mathrm{point}\mathrm{D}\mathrm{on}\stackrel{⌢}{\mathrm{AB}}\mathrm{the}\mathrm{quadrilateral}$$$$\mathrm{ABDCremains}\mathrm{no}\mathrm{longer}\mathbf{convex}.$$$$$$

Commented by RasheedSoomro last updated on 22/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17

$$ifwetakeDonAB,relation:DA=DB+DC$$$$isnotvalid.$$$$thisquistionjustsaythat:$$$$ifDbeapointoncircumcircle,and$$$$weconnectthistovertesesofABC,$$$$lengthoflongestlinefromDtoA,B,or$$$$C,equailstosumofother2lines,$$$$thathavesmallerlengths.$$

Commented by RasheedSoomro last updated on 22/Jun/17

$$\mathrm{Now}\mathrm{the}\mathrm{statement}\mathrm{of}\mathrm{the}\mathrm{theorm}$$$$\mathrm{is}\mathrm{right}\mathrm{sir}!\mathrm{Thanks}!$$

Answered by Tinkutara last updated on 22/Jun/17

$$\mathrm{Since}ABDC\mathrm{is}\mathrm{cyclic},\mathrm{so}$$$$AB.CD+AC.BD=AD.BC$$$$(\mathrm{See}\mathrm{Q}.16070)$$$$\mathrm{Also},AB=BC=AC$$$$\therefore \mathit{C}\mathit{D}+\mathit{B}\mathit{D}=\mathit{A}\mathit{D}$$

Answered by ajfour last updated on 22/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17

$$thankyoumrAjfour.$$

Commented by ajfour last updated on 22/Jun/17

$$LetuscallCD=a,BD=b$$$$In△ADB,$$$$x^2+b^2-s^2=2bx\cos \mathrm{\angle }ADB$$$$\Rightarrow x^2+b^2-s^2=bx.....\left(i\right)$$$$In△CDB,$$$$a^2+b^2-s^2=2ab\cos \mathrm{\angle }CDB$$$$a^2+b^2-s^2=-ab(as\mathrm{\angle }CDB=2\pi /3)$$$$subtractingthieeqn.from\left(i\right):$$$$x^2-a^2=bx+ab$$$$(x-a)(x+a)=b(x+a)$$$$Eitherx=-a,or$$$$x-a=b\Rightarrow x=a+b$$$$\mathit{D}\mathit{A}=\mathit{C}\mathit{D}+\mathit{B}\mathit{D}.$$$$$$

Answered by mrW1 last updated on 22/Jun/17

Commented by b.e.h.i.8.3.4.1.7@gmail.com last updated on 22/Jun/17

$$thankyoumrW1.$$

Commented by mrW1 last updated on 22/Jun/17

$$\mathrm{add}\mathrm{EB}//\mathrm{CD}$$$$\mathrm{\angle }\mathrm{CDA}=\mathrm{\angle }\mathrm{CEB}=60°$$$$...$$$$\mathrm{CDFE}\mathrm{is}\mathrm{parallelogram},$$$$\Rightarrow \mathrm{CD}=\mathrm{CF}$$$$\mathrm{\Delta }\mathrm{DBF}\mathrm{and}\mathrm{\Delta }\mathrm{AFE}\mathrm{are}\mathrm{equilateral},$$$$\Rightarrow \mathrm{DB}=\mathrm{DF}$$$$\Rightarrow \mathrm{CD}=\mathrm{EF}=\mathrm{FA}$$$$\Rightarrow \mathrm{DB}+\mathrm{CD}=\mathrm{DF}+\mathrm{FA}=\mathrm{DA}$$

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