CD and GH are respectively the bisectors of ACB and EGF such that D and H lie on sides AB and FE of ABC and EFG respectively. If ABC ~ FEG, show that: (i) CD AC GH FG (ii) DCB ~ HGE (iii) DCA ~ HGF Fig. 6.36 Fig. 6.37 Fig. 6.38 Fig. 6.39 97
Step-by-Step Solution
Key Concept: Use the Angle‑Bisector Theorem together with the given similarity of the two triangles. The theorem gives the ratio in which a bisector divides the opposite side. Because the corresponding sides of the similar triangles are proportional, the ratios obtained from the two bisectors become equal, leading to the required proportionality and the similarity of the smaller triangles.
Given
- ΔABC \(\sim\) ΔFEG.
- CD bisects \(\angle ACB\) and meets AB at D.
- GH bisects \(\angle EGF\) and meets FE at H.
Step 1 – Write the proportion from the similarity of ΔABC and ΔFEG
Since the triangles are similar, the corresponding sides are in the same ratio. Taking the order \(A\leftrightarrow F,\; B\leftrightarrow E,\; C\leftrightarrow G\), we have
$$\frac{AB}{FE}=\frac{BC}{EG}=\frac{AC}{FG}=k \quad (k>0).$$
Hence
$$\frac{AC}{BC}=\frac{FG}{EG}\qquad\text{and}\qquad \frac{AB}{FE}=k.\tag{1}$$
Step 2 – Apply the Angle‑Bisector Theorem in ΔABC
In ΔABC, CD is the bisector of \(\angle ACB\); therefore
$$\frac{AD}{DB}=\frac{AC}{BC}.\tag{2}$$
Step 3 – Apply the Angle‑Bisector Theorem in ΔFEG
In ΔFEG, GH is the bisector of \(\angle EGF\); therefore
$$\frac{FH}{HE}=\frac{FG}{GE}.\tag{3}$$
Step 4 – Relate the two ratios
From (2) and (3) and using (1) we obtain
$$\frac{AD}{DB}=\frac{AC}{BC}=\frac{FG}{GE}=\frac{FH}{HE}.$$
Thus the two bisectors divide the opposite sides in the same ratio.
Step 5 – Prove (i) \(\displaystyle \frac{CD}{AC}=\frac{GH}{FG}\)
Consider triangles \(\triangle ACD\) and \(\triangle FCG\). They share the angle at C and have the side‑ratio
$$\frac{AD}{DB}=\frac{FH}{HE}$$
from Step 4. By the SAS similarity criterion (two sides in proportion and the included angle equal) we get
$$\frac{CD}{AC}=\frac{GH}{FG}.$$
Hence (i) is proved.
Step 6 – Prove (ii) \(\triangle DCB \sim \triangle HGE\)
From Step 4 we have
$$\frac{DB}{AD}=\frac{HE}{FH}.$$
Using the similarity of the large triangles (1) we also have
$$\frac{BC}{AC}=\frac{GE}{FG}.$$
Now in triangles \(\triangle DCB\) and \(\triangle HGE\):
- \(\angle DCB = \angle HGE\) because they are respectively the halves of the equal angles \(\angle ACB\) and \(\angle EGF\).
- The ratios of the adjacent sides are equal:
$$\frac{DB}{BC}=\frac{HE}{GE}$$
(obtained by dividing the two equalities above).
Thus by the SAS similarity criterion, \(\triangle DCB \sim \triangle HGE\). Hence (ii) is established.
Step 7 – Prove (iii) \(\triangle DCA \sim \triangle HGF\)
Analogous to Step 6, using the equalities
$$\frac{AD}{AC}=\frac{FH}{FG}$$
(from the Angle‑Bisector Theorem and similarity) and the fact that \(\angle DAC = \angle FHG\) (each is the complement of the equal angles at C and G), we obtain the SAS condition for triangles \(\triangle DCA\) and \(\triangle HGF\). Therefore they are similar, proving (iii).
Conclusion
All three required statements follow directly from the Angle‑Bisector Theorem together with the given similarity of the original triangles.
Correct Answer: (i) \(\displaystyle \frac{CD}{AC}=\frac{GH}{FG}\)
(ii) \(\triangle DCB \sim \triangle HGE\)
(iii) \(\triangle DCA \sim \triangle HGF\)