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-

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?

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

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

$$\frac{y}{4}=\frac{x}{z}\Rightarrow yz=4x$$$$6^{2}y+6^{2}z=(y+z)(4^2+yz)$$$$6^2=4^2+4x\Rightarrow x=5$$$$BD=4+5=9✓$$

Commented by efronzo1 last updated on 03/Jul/24

$$\left(\mathrm{i}\right)\underset{―}{⋮}\underbrace{\mathfrak{D}\mathrm{♠}\mathbf{Z}⪅\mathit{\nu }}$$

Commented by dr1001sa last updated on 03/Jul/24

$$whichtheoremuuse$$

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

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

$$\cos \theta =\frac{n^2+d^2-b^2}{2nd}$$$$\cos (\pi -\theta )=\frac{m^2+d^2-a^2}{2md}=-\cos \theta$$$$\frac{m^2+d^2-a^2}{2md}=-\frac{n^2+d^2-b^2}{2nd}$$$$\Rightarrow {mb}^2+{na}^2=(m+n)(d^2+mn)$$$$thisisalsocalled{Stewart}^{\prime }stheorem.$$

Commented by dr1001sa last updated on 03/Jul/24

$$thankyousomuch$$

Answered by A5T last updated on 02/Jul/24

Commented by A5T last updated on 02/Jul/24

$$ConsidercirclecenteredatBwithradius4$$$$\Rightarrow AE\times AF=AS\times AG=AS\times SC=BS\times SD$$$$AE=AB-BE=6-4=2;AF=AE+2\times 4=10$$$$\Rightarrow AS\times AG=2\times 10=20=BS\times SD=4\times SD$$$$\Rightarrow SD=\frac{20}{4}=5\Rightarrow BD=BS+SD=5+4=9$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com