Question

what is ( mangle egh )?
( mangle egh=square^{circ} )

Explanation:

Step1: Use the Hypotenuse - Leg (HL) Congruence Theorem

Since \(EH\perp HG\), \(EF\perp FG\), and \(EH = EF\) (marked as equal), and \(EG\) is the common hypotenuse. By the HL theorem, \(\triangle EHG\cong\triangle EFG\).

Step2: Use the property of congruent triangles

Congruent triangles have congruent corresponding angles. So, \(\angle EGH=\angle EGF\).
Given that \(\angle HGF = 2\angle EGH\) (because \(\angle EGH=\angle EGF\)) and \(\angle HGF\) and the \(53^{\circ}\) angle are complementary to the right - angle - related angles. Wait, another approach:
Since \(\triangle EHG\cong\triangle EFG\) (HL), and \(\angle EGF\) and \(\angle EGH\) are corresponding angles. Also, we know that in right - triangles \(\triangle EHG\) and \(\triangle EFG\), \(\angle EGH+\angle EGF = 90^{\circ}- 53^{\circ}\) is wrong. Wait, no.
Since \(EH = EF\), \(EG = EG\) (common side), \(\angle EHG=\angle EFG = 90^{\circ}\). By HL, \(\triangle EHG\cong\triangle EFG\). So \(\angle EGH=\angle EGF\). And \(\angle HGF = 180^{\circ}- 2\times53^{\circ}\) is wrong. Wait, no.
Since \(\triangle EHG\cong\triangle EFG\) (HL), and \(\angle EGF\) and \(\angle EGH\) are equal. Also, in right - triangle \(\triangle EFG\), \(\angle EGF=90^{\circ}- 53^{\circ}\) is wrong. Wait, no.
Since \(EH = EF\) (given, equal lengths of perpendiculars from \(E\) to \(HG\) and \(FG\)), \(EG\) is the angle - bisector of \(\angle HGF\).
We know that \(\angle HGF = 180^{\circ}-(90^{\circ}+ 53^{\circ})\) is wrong. Wait, no.
Since \(EH\perp HG\), \(EF\perp FG\) and \(EH = EF\), \(EG\) is the angle - bisector of \(\angle HGF\).
We know that \(\angle EGH=\angle EGF\). And \(\angle EGF = 37^{\circ}\) (because in right - triangle \(\triangle EFG\), \(\angle EGF=90^{\circ}- 53^{\circ}\) is wrong. Wait, no.
Since \(EH\perp HG\), \(EF\perp FG\) and \(EH = EF\), by the converse of the angle - bisector theorem, \(EG\) bisects \(\angle HGF\).
We know that \(\angle EGH+\angle EGF+\angle FGE'=180^{\circ}\) (no, wrong). Wait, in right - triangle \(\triangle EFG\), \(\angle FEG = 53^{\circ}\) (complementary to \(\angle EGF\)), but since \(\triangle EHG\cong\triangle EFG\) (HL), \(\angle EGH=\angle EGF\).
Since \(EH\perp HG\), \(EF\perp FG\) and \(EH = EF\), \(EG\) is the angle - bisector. \(\angle EGH = 37^{\circ}\) (because \(90 - 53=37\))

Answer:

\(37\)