Question

question
given: (abcd) is a parallelogram, (overline{bg} cong overline{dh}) and (overline{ae} cong overline{fc}).
prove: (overline{ge} cong overline{hf}).

step | statement | reason
1 | (abcd) is a parallelogram, (overline{bg} cong overline{dh}), (overline{ae} cong overline{fc}) | given
2 | (overline{ef} cong overline{ef}) | reflexive property
3 | (overline{af} cong overline{ce}) | congruent segments added to congruent segments form congruent segments
4 | (overline{ad} cong overline{bc}) | opposite sides of a parallelogram / rhombus / rectangle / square are congruent
5 | (overline{ah} cong overline{cg}) | select a reason...

(image of parallelogram (abcd) with points (g) on (bc), (h) on (ad), (e) and (f) on (ac); note: (overline{ac}), (overline{bc}) and (overline{ad}) are congruent.)

Explanation:

Step1: Recall properties of parallelograms

In a parallelogram \(ABCD\), \(AD \cong BC\) (opposite sides of a parallelogram are congruent, as given in step 4).

Step2: Use given congruent segments

We know \(BG \cong DH\) (given). Let's express \(AH\) and \(CG\) in terms of these segments. Since \(AD = AH + HD\) and \(BC = BG + GC\), and \(AD \cong BC\), \(BG \cong DH\), we can substitute \(HD\) with \(BG\) in the equation for \(AD\). So, \(AH + BG= BG + CG\). Subtracting \(BG\) from both sides (subtraction property of congruence, or more precisely, congruent segments subtracted from congruent segments form congruent segments), we get \(AH \cong CG\). The reason is: Congruent segments subtracted from congruent segments form congruent segments (since \(AD \cong BC\) and \(DH \cong BG\), then \(AD - DH \cong BC - BG\), which is \(AH \cong CG\)).

Step3: Prove triangles congruent (later steps, but for step 5 reason)

For step 5, the reason for \(AH \cong CG\) is that congruent segments (\(AD \cong BC\)) minus congruent segments (\(DH \cong BG\)) are congruent (subtraction property of congruent segments: if \(a \cong b\) and \(c \cong d\), then \(a - c \cong b - d\)).

Now, to continue the proof for \(GE \cong HF\):

Step4: Identify triangles

We can show \(\triangle AGE \cong \triangle DHF\) or \(\triangle GEC \cong \triangle HFA\), but more likely, look at \(\triangle GAE\) and \(\triangle HCF\) or \(\triangle GEF\) and \(\triangle HFE\). Wait, alternatively, since \(AE \cong FC\) (given), \(AH \cong CG\) (just proved), and \(\angle EAH \cong \angle FCG\) (since \(AB \parallel CD\) in parallelogram, so alternate interior angles, or since \(AD \parallel BC\), \(\angle DAB \cong \angle BCD\), and \(AE, FC\) are segments on \(AC\), so \(\angle EAH=\angle FCG\)). Then by SAS (side - angle - side) congruence, \(\triangle AEH \cong \triangle CFG\)? Wait, no, let's get back to step 5. The reason for \(AH \cong CG\) is: Congruent segments (\(AD\) and \(BC\)) minus congruent segments (\(DH\) and \(BG\)) are congruent (i.e., \(AD - DH \cong BC - BG\) implies \(AH \cong CG\)).

After proving \(AH \cong CG\), \(AE \cong FC\) (given), and \(\angle EAH \cong \angle FCG\) (alternate interior angles because \(AD \parallel BC\) in parallelogram \(ABCD\) and \(AC\) is a transversal), then by SAS, \(\triangle AHE \cong \triangle CGF\). Then \(HE \cong GF\), and with \(EF \cong EF\) (reflexive), by SSS, \(\triangle GEF \cong \triangle HFE\), so \(GE \cong HF\). But for step 5, the reason is "Congruent segments subtracted from congruent segments form congruent segments" (because \(AD \cong BC\) and \(DH \cong BG\), so \(AD - DH \cong BC - BG\) which is \(AH \cong CG\)).

Answer:

The reason for step 5 (\(AH \cong CG\)) is "Congruent segments subtracted from congruent segments form congruent segments" (since \(AD \cong BC\) and \(DH \cong BG\), so \(AD - DH \cong BC - BG\), hence \(AH \cong CG\)). And to prove \(GE \cong HF\), we can use triangle congruence (SAS or SSS) after establishing the necessary congruent sides and angles. The key step for \(AH \cong CG\) uses the property that if two congruent segments (\(AD, BC\)) have two other congruent segments (\(DH, BG\)) subtracted from them, the results are congruent.