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Problem 370

# $$\underline{E B A}$$ and $$\underline{E G D}$$ are secants of $$\odot P$$. Chords $$\underline{A F}$$ intersects $$\underline{E G D}$$ at point $$G$$. If $E B=5, B A=7, E C=4, G D=3$$, and$$A G=6$$, find$$\mathrm{GF}$

Expert verified
The length of GF is $$5$$ units.
See the step by step solution

## Step 1: 1. Power Theorem of Secants (S-S)

We will use the secant-secant Power Theorem, which states that for any two secant segments intersecting outside a circle, the product of the lengths of the outer segment and the whole secant segment for one secant equals the product of the lengths of the outer segment and the whole secant segment for the other secant. In this case, we will apply the theorem to the secants EA and EG: - Secant EA: Outer segment - EB, Length: 5 Whole secant - EA, Length: EB + BA, Length: 5 + 7 = 12 - Secant EG: Outer segment - EC, Length: 4 Whole secant - EG, Length: EC + CG, Length: 4 + CG (where CG needs to be determined)

## Step 2: 2. Power Theorem Formula

Let's plug in the lengths of the given segments into the Power Theorem formula: $EB * EA = EC * EG$

## Step 3: 3. Substituting given values

Now substitute the given values and solve for CG: $$5 * 12 = 4 * (4 + CG)$$

## Step 4: 4. Solve for the length of CG

Solve for CG: $$60 = 16 + 4CG$$ $$44 = 4CG$$ $$CG = 11$$

## Step 5: 5. Triangle AGF lengths

Recall that AF intersects EG at point G, and AG = 6 (given). Also, CG is a part of EG. Now we know the length of CG, which is 11, we can find the length of GF. Since AF is a straight line: GF = CG - AG GF = 11 - 6

## Step 6: 6. Calculate the length of GF

Now solve for GF: GF = 5 So the length of GF is 5 units.

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