In \( \triangle G H I, \overline{I G} \cong \overline{H I} \) and \( \mathrm{m} \angle H=51^{\circ} \). Find \( \mathrm{m} \angle G \)

In triangle \( GHI \), we know that \( IG \cong HI \), which means that triangle \( GHI \) is isosceles with the two sides \( IG \) and \( HI \) being equal. In an isosceles triangle, the angles opposite the equal sides are also equal. Therefore, we have: \[ m \angle G = m \angle I \] Given that the sum of the angles in any triangle is always \( 180^\circ \), we can set up the following equation based on the angles in triangle \( GHI \): \[ m \angle G + m \angle H + m \angle I = 180^\circ \] Substituting \( m \angle H = 51^\circ \) and the fact that \( m \angle G = m \angle I \) (let's denote them as \( x \)), we have: \[ x + 51^\circ + x = 180^\circ \] This simplifies to: \[ 2x + 51^\circ = 180^\circ \] Now, subtract \( 51^\circ \) from both sides: \[ 2x = 180^\circ - 51^\circ \] \[ 2x = 129^\circ \] Next, divide by 2: \[ x = \frac{129^\circ}{2} = 64.5^\circ \] So, we find that: \[ m \angle G = 64.5^\circ \] Thus, the measure of angle \( G \) is \( \boxed{64.5^\circ} \).