Question

given: $angle fcongangle j$, $overline{fd}perpoverline{dh}$, $overline{jh}perpoverline{dh}$, $overline{fg}congoverline{jg}$
prove: $g$ is the mid - point of $overline{dh}$.

1. $\triangle fdgcong\triangle jhg$
2. $angle fgdcongangle jhg$
3. $mangle f + mangle fgd=mangle fdg$

Explanation:

Step1: Recall triangle congruence

Given \(\angle F\cong\angle J\), \(\overline{FD}\perp\overline{DH}\), \(\overline{JH}\perp\overline{DH}\) (so \(\angle FDG=\angle JHG = 90^{\circ}\)) and \(\overline{FG}\cong\overline{JG}\), we use AAS to prove \(\triangle FDG\cong\triangle JHG\).

Step2: Apply CPCTC

Since \(\triangle FDG\cong\triangle JHG\), corresponding sides are congruent. So \(DG = HG\).

Step3: Define mid - point

A mid - point of a line segment divides it into two equal parts. Since \(DG = HG\), \(G\) is the mid - point of \(\overline{DH}\).

Answer:

The statement in step 7 is incorrect. In a triangle, the sum of two non - exterior angles is equal to the exterior angle. In \(\triangle FDG\), \(m\angle F + m\angle FGD=m\angle FDH\) (exterior - angle property of a triangle), not \(m\angle FDG\). The correct reasoning for step 7 to prove \(G\) is the mid - point of \(\overline{DH}\) after having \(\triangle FDG\cong\triangle JHG\) is:

1. Since \(\triangle FDG\cong\triangle JHG\) by Angle - Angle - Side (AAS) congruence criterion.
2. By the Corresponding Parts of Congruent Triangles are Congruent (CPCTC), we have \(DG = HG\).
3. By the definition of a mid - point (a point that divides a line segment into two equal parts), \(G\) is the mid - point of \(\overline{DH}\).