A circle is shown. Chords A C and B D intersect at point E. The length of A E is x, the length of E C is x + 12, the length of B E is x + 2, and the length of E D is x + 5. BE is 2 units longer than AE, DE is 5 units longer than AE, and CE is 12 units longer than AE

Complete Question:

A circle is shown. Chords A C and B D intersect at point E. The length of A E is x, the length of E C is x + 12, the length of B E is x + 2, and the length of E D is x + 5. 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?

*The circle is attached below in the attachment

BD = 11

Step-by-step explanation:

Given:

AE = x

EC = x + 12

BE = x + 2

ED = x + 5

Required:

Length of BD

Solution:

First, create an equation to find the value of (x + 2)(x + 5) = (x)(x + 12) => intersecting chords theorem

x(x + 5) + 2(x + 5) = x(x + 12)

x² + 5x + 2x + 10 = x² + 12x

Add like terms

x² + 7x + 10 = x² + 12x

Subtract x² from each side

x² + 7x + 10 - x² = x² + 12x - x²

7x + 10 = 12x

10 = 12x - 7x

10 = 5x

10/5 = 5x/5

2 = x

x = 2

Next, find BD:

BD = BE + ED

BD = x + 2 + x + 5

Add like terms

BD = 2x + 7

Plug in the value of x

BD = 2(2) + 7

BD = 4 + 7

BD = 11

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step-by-step explanation: