CD and GH are respectively the bisectors of \(\angle ACB\) and \(\angle EGF\) such that D and H lie on sides AB and FE of \(\triangle ABC\) and \(\triangle EFG\) respectively. If \(\triangle ABC \sim \triangle FEG\), show that:
(i) \(\dfrac{CD}{GH} = \dfrac{AC}{FG}\)
(ii) \(\triangle DCB \sim \triangle HGE\)
(iii) \(\triangle DCA \sim \triangle HGF\)
Given: \(\triangle ABC \sim \triangle FEG\). CD and GH are internal bisectors of \(\angle ACB\) and \(\angle EGF\) respectively, with D on AB and H on FE.
From the similarity \(\triangle ABC \sim \triangle FEG\), we have the correspondence:
\[A \leftrightarrow F,\quad B \leftrightarrow E,\quad C \leftrightarrow G\]
and so
\[\angle ACB = \angle FGE,\quad \angle ABC = \angle FEG,\quad \angle BAC = \angle GFE.\]
(i) Prove \(\dfrac{CD}{GH} = \dfrac{AC}{FG}\)
Because similarity is a rigid “shape-preserving” scaling, the angle bisector at vertex C of \(\triangle ABC\) corresponds to the angle bisector at vertex G of \(\triangle FEG\). Thus segment CD in the first triangle corresponds to segment GH in the second triangle, just as side AC corresponds to side FG.
Therefore, the ratio of corresponding bisectors equals the ratio of any pair of corresponding sides:
\[\frac{CD}{GH} = \frac{AC}{FG}.\]
(ii) Prove \(\triangle DCB \sim \triangle HGE\)
Since CD and GH are angle bisectors,
\[\angle ACD = \angle DCB = \tfrac{1}{2}\angle ACB,\]
\[\angle FGH = \angle HGE = \tfrac{1}{2}\angle FGE.\]
But \(\angle ACB = \angle FGE\) (from similarity of the big triangles). Hence,
\[\angle DCB = \angle HGE.\tag{1}\]
Also, D lies on AB and H lies on FE. Using \(\triangle ABC \sim \triangle FEG\), we have
\[\angle ABC = \angle FEG.\]
Angles at B and E made with the same sides give
\[\angle DBC = \angle HEG.\tag{2}\]
From (1) and (2), two angles of \(\triangle DCB\) equal two angles of \(\triangle HGE\). Therefore, by AA similarity,
\[\triangle DCB \sim \triangle HGE.\]
(iii) Prove \(\triangle DCA \sim \triangle HGF\)
Again using the angle bisectors,
\[\angle DCA = \angle ACD = \tfrac{1}{2}\angle ACB,\]
\[\angle HGF = \angle FGH = \tfrac{1}{2}\angle FGE.\]
As before, \(\angle ACB = \angle FGE\), so
\[\angle DCA = \angle HGF.\tag{3}\]
From the similarity of the big triangles,
\[\angle BAC = \angle GFE.\]
D lies on AB and H lies on FE, so angle at A in \(\triangle DCA\) equals angle at F in \(\triangle HGF\):
\[\angle DAC = \angle HFG.\tag{4}\]
Using (3) and (4), two angles of \(\triangle DCA\) equal two angles of \(\triangle HGF\). Thus, by AA similarity,
\[\triangle DCA \sim \triangle HGF.\]
Hence all three statements (i), (ii) and (iii) are proved.