Question

gh = ab = fg = m<a = m<e = m<g = m<b =

Explanation:

Step1: Identify congruent sides

Since the two figures are congruent (assuming they are congruent trapezoids), \(GH = DC = 6\), \(AB=EF\) (but from the left - hand figure \(AB\) can be found using the property of congruent trapezoids. Wait, actually, if we consider the sum of interior angles of a trapezoid. But first, for side lengths:
For \(AB\):
The sum of the lengths of non - parallel sides and parallel sides. Wait, no, since the two trapezoids are congruent. In the left - hand trapezoid \(ABCD\) (right - angled trapezoid), and the right - hand trapezoid \(EFGH\).
\(GH = DC = 6\) (corresponding sides of congruent figures).
\(AB\): In trapezoid \(ABCD\), using the property of congruent trapezoids (if they are congruent). Wait, actually, if we assume the two trapezoids are congruent.
\(AB = 15\) (corresponding to the vertical non - parallel side of the other trapezoid).
\(FG=5\) (corresponding to \(CB\) in the left - hand trapezoid).

Step2: Calculate angles

The sum of interior angles of a quadrilateral is \(360^{\circ}\). In trapezoid \(ABCD\) (right - angled trapezoid, \(\angle C=\angle B = 90^{\circ}\)).
Let \(\angle A=x\) and \(\angle D\).
In trapezoid \(EFGH\), \(\angle H = 110^{\circ}\), \(\angle F=\angle C=\angle B = 90^{\circ}\), \(\angle G=\angle C = 90^{\circ}\) (right - angled trapezoid property, if congruent).
For \(\angle A\):
Since the sum of interior angles of a quadrilateral \(ABCD\) is \(360^{\circ}\), \(\angle A+\angle B+\angle C+\angle D=360^{\circ}\). In trapezoid \(EFGH\), \(\angle H = 110^{\circ}\), \(\angle F = 90^{\circ}\), \(\angle G=90^{\circ}\), \(\angle E\).
Since the trapezoids are congruent, \(\angle A=\angle E\).
Using the sum of interior angles formula for a quadrilateral \(S=(n - 2)\times180^{\circ}\) (where \(n = 4\), \(S = 360^{\circ}\)).
In trapezoid \(ABCD\): \(\angle A+\angle B+\angle C+\angle D=360^{\circ}\). If we assume \(\angle D=\angle H = 110^{\circ}\) (corresponding angles of congruent trapezoids), \(\angle B=\angle F=\angle C=\angle G = 90^{\circ}\).
\(\angle A=360-(90 + 90+110)=70^{\circ}\).
\(\angle E=\angle A = 70^{\circ}\), \(\angle G = 90^{\circ}\), \(\angle B=90^{\circ}\)

Answer:

\(GH = 6\), \(AB = 15\), \(FG = 5\), \(m\angle A=70^{\circ}\), \(m\angle E = 70^{\circ}\), \(m\angle G=90^{\circ}\), \(m\angle B=90^{\circ}\)