Question

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

Explanation:

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

If the hypotenuse and a leg of one right - triangle are congruent to the hypotenuse and a leg of another right - triangle, then the two right - triangles are congruent. Here, in right - triangles \( \triangle FEH\) and \( \triangle FGH\), \(FH = FH\) (common hypotenuse), \(EH=GH = 95\). So, \( \triangle FEH\cong\triangle FGH\) by the HL theorem.

Step2: Use the property of congruent triangles

Since \( \triangle FEH\cong\triangle FGH\), then \( \angle EFH=\angle GFH\).

Step3: Find the measure of \( \angle EFH\)

We know that \( \angle EFG=\angle EFH+\angle GFH\). Given \( \angle E = 90^{\circ}\), \( \angle G=90^{\circ}\), and the sum of angles around a point is not needed. But since \( \triangle FEH\cong\triangle FGH\), and if we assume \( \angle EFG\) is composed of \( \angle EFH\) and \( \angle GFH\). Let \(x = \angle EFH=\angle GFH\). We know that \(x + 50^{\circ}+x=180^{\circ}- 90^{\circ}-90^{\circ}+ 180^{\circ}\) (using the fact that the non - right angles of congruent right - triangles). But a simpler way: since \( \triangle FEH\cong\triangle FGH\), and \( \angle EFG\) (the angle at \(F\) for the two congruent right - triangles) has \( \angle EFH\) and \( \angle GFH\) as parts. Since the triangles are congruent, \( \angle EFH = 40^{\circ}\) (because \(180^{\circ}-90^{\circ}-50^{\circ}=40^{\circ}\) for the non - congruent angle part. Wait, another approach:
The sum of angles in a quadrilateral \(EFGH\) (but \(E\) and \(G\) are right angles). The sum of interior angles of a quadrilateral is \((4 - 2)\times180^{\circ}=360^{\circ}\). Let \( \angle EFH=x\) and \( \angle GFH = x\) (from congruent triangles). Then \(90^{\circ}+90^{\circ}+x + 50^{\circ}+x=360^{\circ}\).

$$2x=360^{\circ}-(90^{\circ}+90^{\circ}+50^{\circ})$$

$$2x = 130^{\circ}$$

(incorrect). Wait, correct approach:
Since \( \triangle FEH\cong\triangle FGH\) (HL: \(EH = GH = 95\), \(FH=FH\)), then \( \angle EFH=\angle GFH\). And \( \angle EFG\) (the angle at \(F\) for the two right - triangles) has \( \angle EFG=180^{\circ}- 90^{\circ}-50^{\circ}\) (using the fact that the line \(FG\) and \(FE\) with the right - angles. Wait, no.
We know that \( \angle EFG\) (the angle we want to find \( \angle EFH\)):
Since \( \triangle FEH\cong\triangle FGH\) (HL), and \( \angle EFG\) (the angle at \(F\)):
\( \angle EFH+\angle GFH = 180^{\circ}- 90^{\circ}-50^{\circ}\) (no, wrong. Correct:
In right - triangle \( \triangle FGH\), \( \angle GFH=180^{\circ}-90^{\circ}-50^{\circ}=40^{\circ}\). Since \( \triangle FEH\cong\triangle FGH\), \( \angle EFH=\angle GFH = 40^{\circ}\)

Answer:

\(40\)