Post on 17-Dec-2015
The proof…
Then we construct line DJ which is parallel to CI and GH. Let M be the point where CG and DJ intersect.
(image with line DJ)
The proof…
Notice that angle BCD and angle MCI
are both 90°
therefore angle BCG and angle DCI are congruent.
The proof…
Moreover, sides BC and CD are congruent because they are the sides of the same square ABCD.
The proof…
Knowing that angles BCG and DCI are congruent, and sides BC and CD are also congruent, therefore sides BG and DI must be congruent as well.
By SAS, ΔCBG and ΔCDI are congruent.
The proof…
By SAS, ΔCBG and ΔCDI are congruent.
So,
½ (BC)(CD) ≡ ½ (CI)(CM)
(CD)(CD)≡(CI)(CM)…(since BC=CD)
(CD) 2 ≡(CI)(CM)
Notice that (CD) 2 = area of ABCD
And (CI)(CM)= area of CMJI
Therefore, the area of ABCD is equal to the area of the CMJI.
The proof…
Notice that angle DGF and angle MGH
are both 90°
Therefore angle BCG and angle DCI
are congruent.
The proof…
Moreover, sides FG and DG are congruent because they are
the sides of the same square
ABCD.
The proof…
Knowing that angles FGC and DGH are congruent, and sides FG and DG are also congruent, therefore sides FC and DH must be congruent as well.
(image of the two congruent triangles)
By SAS, ΔGFC and ΔGDH are congruent.
The proof…
By SAS, ΔGFC and ΔGDH are congruent.
So,
½ (FG)(DG) ≡ ½ (GH)(GM)
(DG)(DG)≡(GH)(GM)…(since FG=DG)
(DG) 2 ≡(GH)(GM)
Notice that (DG) 2 = area of DEFG
And (GH)(GM)= area of GMJH
Therefore, the area of DEFG is equal to the area of the GMJH.
The proof…
FINALLY we merge what we got from Part 1 and Part 2.
Using some algebra we get:
(CD)2 + (DG)2 =(CI)(CM) + (GH) (GM)
(CD)2 + (DG)2 =(CI)(CM) + (CI) (GM) …since GH=CI
(CD)2 + (DG)2 =(CI)(CM + GM)
(CD)2 + (DG)2 =(CI)(CG)…since CM+GM=CG
(CD)2 + (DG)2 =(CG)(CG)…since CI=CG
(CD)2 + (DG)2 =(CG)2