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Question Number 209162 by dr1001sa last updated on 02/Jul/24

Cyclic quadrilateral ABCD is inscribed in circle. Point S is intersection of diagonals AC and BD (S is not center of the circle). If AB=BC=6 and BS=4, what is length of BD?

$$ \\ $$Cyclic quadrilateral ABCD is inscribed in circle. Point S is intersection of diagonals AC and BD (S is not center of the circle). If AB=BC=6 and BS=4, what is length of BD?

Answered by mr W last updated on 02/Jul/24

Commented by mr W last updated on 03/Jul/24

(y/4)=(x/z) ⇒yz=4x 6^2 y+6^2 z=(y+z)(4^2 +yz) 6^2 =4^2 +4x ⇒x=5 BD=4+5=9 ✓

$$\frac{{y}}{\mathrm{4}}=\frac{{x}}{{z}}\:\Rightarrow{yz}=\mathrm{4}{x} \\ $$$$\mathrm{6}^{\mathrm{2}} {y}+\mathrm{6}^{\mathrm{2}} {z}=\left({y}+{z}\right)\left(\mathrm{4}^{\mathrm{2}} +{yz}\right) \\ $$$$\mathrm{6}^{\mathrm{2}} =\mathrm{4}^{\mathrm{2}} +\mathrm{4}{x}\:\Rightarrow{x}=\mathrm{5}\: \\ $$$${BD}=\mathrm{4}+\mathrm{5}=\mathrm{9}\:\checkmark \\ $$

Commented by efronzo1 last updated on 03/Jul/24

(i) ⋮D♠Z⪅𝛎

$$\:\:\left(\mathrm{i}\right)\:\:\underline{\vdots}\underbrace{\mathfrak{D}\spadesuit\boldsymbol{\mathrm{Z}}\lessapprox\boldsymbol{\nu}} \\ $$

Commented by dr1001sa last updated on 03/Jul/24

which theorem u use

$${which}\:{theorem}\:{u}\:{use} \\ $$

Commented by mr W last updated on 03/Jul/24

Commented by mr W last updated on 03/Jul/24

cos θ=((n^2 +d^2 −b^2 )/(2nd)) cos (π−θ)=((m^2 +d^2 −a^2 )/(2md))=−cos θ ((m^2 +d^2 −a^2 )/(2md))=−((n^2 +d^2 −b^2 )/(2nd)) ⇒ mb^2 +na^2 =(m+n)(d^2 +mn) this is also called Stewart′s theorem.

$$\mathrm{cos}\:\theta=\frac{{n}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{nd}} \\ $$$$\mathrm{cos}\:\left(\pi−\theta\right)=\frac{{m}^{\mathrm{2}} +{d}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{md}}=−\mathrm{cos}\:\theta \\ $$$$\frac{{m}^{\mathrm{2}} +{d}^{\mathrm{2}} −{a}^{\mathrm{2}} }{\mathrm{2}{md}}=−\frac{{n}^{\mathrm{2}} +{d}^{\mathrm{2}} −{b}^{\mathrm{2}} }{\mathrm{2}{nd}} \\ $$$$\Rightarrow\:{mb}^{\mathrm{2}} +{na}^{\mathrm{2}} =\left({m}+{n}\right)\left({d}^{\mathrm{2}} +{mn}\right) \\ $$$${this}\:{is}\:{also}\:{called}\:{Stewart}'{s}\:{theorem}. \\ $$

Commented by dr1001sa last updated on 03/Jul/24

thank you so much

$${thank}\:{you}\:{so}\:{much} \\ $$

Answered by A5T last updated on 02/Jul/24

Commented by A5T last updated on 02/Jul/24

Consider circle centered at B with radius 4 ⇒AE×AF=AS×AG=AS×SC=BS×SD AE=AB−BE=6−4=2;AF=AE+2×4=10 ⇒AS×AG=2×10=20=BS×SD=4×SD ⇒SD=((20)/4)=5⇒BD=BS+SD=5+4=9

$${Consider}\:{circle}\:{centered}\:{at}\:{B}\:{with}\:{radius}\:\mathrm{4} \\ $$$$\Rightarrow{AE}×{AF}={AS}×{AG}={AS}×{SC}={BS}×{SD} \\ $$$${AE}={AB}−{BE}=\mathrm{6}−\mathrm{4}=\mathrm{2};{AF}={AE}+\mathrm{2}×\mathrm{4}=\mathrm{10} \\ $$$$\Rightarrow{AS}×{AG}=\mathrm{2}×\mathrm{10}=\mathrm{20}={BS}×{SD}=\mathrm{4}×{SD} \\ $$$$\Rightarrow{SD}=\frac{\mathrm{20}}{\mathrm{4}}=\mathrm{5}\Rightarrow{BD}={BS}+{SD}=\mathrm{5}+\mathrm{4}=\mathrm{9} \\ $$

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