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?


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

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

	$\frac{y}{4} = \frac{x}{z} \Rightarrow y z = 4 x$ $6^{2} y + 6^{2} z = (y + z) (4^{2} + y z)$ $6^{2} = 4^{2} + 4 x \Rightarrow x = 5$ $B D = 4 + 5 = 9 ✓$
	Commented by efronzo1 last updated on 03/Jul/24

	$(i) \underset{―}{⋮} \underset{⏟}{D ♠ Z ⪅ ν}$
	Commented by dr1001sa last updated on 03/Jul/24

	$w h i c h t h e o r e m u u s e$
	Commented by mr W last updated on 03/Jul/24

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

	$\cos θ = \frac{n^{2} + d^{2} - b^{2}}{2 n d}$ $\cos (π - θ) = \frac{m^{2} + d^{2} - a^{2}}{2 m d} = - \cos θ$ $\frac{m^{2} + d^{2} - a^{2}}{2 m d} = - \frac{n^{2} + d^{2} - b^{2}}{2 n d}$ $\Rightarrow {m b}^{2} + {n a}^{2} = (m + n) (d^{2} + m n)$ $t h i s i s a l s o c a l l e d {S t e w a r t}^{'} s t h e o r e m .$
	Commented by dr1001sa last updated on 03/Jul/24

	$t h a n k y o u s o m u c h$
	Answered by A5T last updated on 02/Jul/24

	Commented by A5T last updated on 02/Jul/24

	$C o n s i d e r c i r c l e c e n t e r e d a t B w i t h r a d i u s 4$ $\Rightarrow A E \times A F = A S \times A G = A S \times S C = B S \times S D$ $A E = A B - B E = 6 - 4 = 2; A F = A E + 2 \times 4 = 10$ $\Rightarrow A S \times A G = 2 \times 10 = 20 = B S \times S D = 4 \times S D$ $\Rightarrow S D = \frac{20}{4} = 5 \Rightarrow B D = B S + S D = 5 + 4 = 9$
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