QUESTION IMAGE
Question
be is 2 units longer than ae, de is 5 units longer than ae, and ce is 12 units longer than ae. what is bd? units a e c b x + 2 x x + 12 d x + 5
Step1: Apply the Intersecting Chords Theorem
The Intersecting Chords Theorem states that if two chords intersect at a point, then the product of the lengths of the segments of one chord is equal to the product of the lengths of the segments of the other chord. For chords \(AC\) and \(BD\) intersecting at \(E\), we have \(AE \times CE = BE \times DE\).
Given \(AE = x\), \(BE = x + 2\), \(DE = x + 5\), and \(CE = x + 12\). Substituting these into the theorem:
\[
x(x + 12) = (x + 2)(x + 5)
\]
Step2: Expand and Simplify the Equation
Expand both sides:
Left side: \(x^2 + 12x\)
Right side: \(x^2 + 5x + 2x + 10 = x^2 + 7x + 10\)
Now, set them equal:
\[
x^2 + 12x = x^2 + 7x + 10
\]
Subtract \(x^2\) from both sides:
\[
12x = 7x + 10
\]
Subtract \(7x\) from both sides:
\[
5x = 10
\]
Divide both sides by 5:
\[
x = 2
\]
Step3: Find the Lengths of \(BE\) and \(DE\)
We know \(BE = x + 2\) and \(DE = x + 5\). Substitute \(x = 2\):
\(BE = 2 + 2 = 4\)
\(DE = 2 + 5 = 7\)
Step4: Calculate the Length of \(BD\)
Since \(BD = BE + DE\) (because \(E\) is between \(B\) and \(D\)):
\[
BD = 4 + 7 = 11
\]
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