Question

if $\angle efh \cong \angle gfh$, $eh = s + 6$, and $gh = 2s$, what is $gh$?
diagram: triangle with vertices h, g, f, e; right angles at g (between gf and gh) and e (between ef and eh); fh is a common segment, and $\angle efh \cong \angle gfh$
$gh = \square$
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Explanation:

Step1: Identify congruent segments

Since $\angle EFH \cong \angle GFH$, and $FG \perp GH$, $FE \perp EH$ (right angles), by the Angle - Bisector Theorem (a point on the angle bisector is equidistant from the sides of the angle), we have $EH = GH$.

Step2: Set up the equation

We know that $EH=s + 6$ and $GH = 2s$. So we set up the equation $s+6=2s$.

Step3: Solve for s

Subtract $s$ from both sides of the equation $s + 6=2s$. We get $6=2s - s$, which simplifies to $s = 6$.

Step4: Find the length of GH

Since $GH = 2s$ and $s = 6$, substitute $s$ into the formula for $GH$. So $GH=2\times6 = 12$.

Answer:

12