Question

\[CD\] and \[GH\] are respectively the bisectors of \[\angle ACB\] and \[\angle EGF\] such that \[D\] and \[H\] lie on sides \[AB\] and \[EF\] of \[\Delta ABC\] and \[\Delta EFG\] respectively. If, \[\Delta ABC \sim \Delta EFG\], show that:
(i). \[\dfrac{{CD}}{{GH}} = \dfrac{{AC}}{{FG}}\]
(ii). \[\Delta DCB \sim \Delta HGE\]
(iii). \[\Delta DCA \sim \Delta HGF\]

Answer 1

Hint: We know that; for any two right-angle triangles, if one of the other two angles is equal then the third angle is also equal to each other. This condition is known as A-A-A similarity condition. Then, we can say that the triangles are similar.
By using the condition of similarity, we can prove the given triangles are similar.
Again, we know that, if two triangles are similar to each other, then the ratio of their corresponding sides is proportional. Using this condition, we can prove the first problem.

Complete step-by-step answer:
It is given that; \[CD\] and \[GH\] are respectively the bisectors of \[\angle ACB\] and \[\angle EGF\] such that \[D\] and \[H\] lie on sides \[AB\] and \[EF\] of \[\Delta ABC\] and \[\Delta EFG\]respectively. Again, \[\Delta ABC\] and \[\Delta EFG\] are similar. Then,
We have to show that,
(i). \[\dfrac{{CD}}{{GH}} = \dfrac{{AC}}{{FG}}\]
(ii). \[\Delta DCB \sim \Delta HGE\]
(iii). \[\Delta DCA \sim \Delta HGF\]
               

We know that the corresponding angles of similar triangles are equal.
Here, \[\Delta ABC\] and \[\Delta EFG\] are similar.
Then, \[\angle ACB = \angle EGF\] as they are corresponding angles of similar triangles.
Since, \[CD\]and \[GH\] are respectively the bisectors of \[\angle ACB\] and \[\angle EGF\] such that \[D\] and \[H\] lie on sides \[AB\] and \[EF\] of \[\Delta ABC\] and \[\Delta EFG\] respectively, then,
\[\angle ACB = 2\angle ACD = 2\angle BCD\]
Similarly, \[\angle EGF = 2\angle FGH = 2\angle HGE\]
So, we have, \[\angle ACD = \angle FGH\] and \[\angle DCB = \angle HGE\] … (1)
Also, \[\angle A = \angle F\] and \[\angle B = \angle E\] …. (2)
In \[\Delta ACD\] and \[\Delta FGH\]
From (2), \[\angle A = \angle F\]
From (1), \[\angle ACD = \angle FGH\]
By A-A-A condition of similarity we get, \[\Delta ACD \sim \Delta FGH\]
We know that, if two triangles are similar to each other, then the ratio of their corresponding sides is proportional.
Here, \[\Delta ACD\] and \[\Delta FGH\] are similar to each other, so, we have,
\[\dfrac{{CD}}{{GH}} = \dfrac{{AC}}{{FG}} = \dfrac{{AD}}{{FH}}\]
From the above condition we get,
\[\dfrac{{CD}}{{GH}} = \dfrac{{AC}}{{FG}}\]
In \[\Delta DCB\] and \[\Delta HGE\]
From (2), \[\angle B = \angle E\]
From (1), \[\angle DCB = \angle HGE\]
By A-A-A condition of similarity we get, \[\Delta DCB \sim \Delta HGE\]
Hence,
\[\dfrac{{CD}}{{GH}} = \dfrac{{AC}}{{FG}}\]
\[\Delta DCB \sim \Delta HGE\]
\[\Delta DCA \sim \Delta HGF\]

Note: If the three angles of any triangle are equal to the respective angles of another triangle, then the triangle is called a similar triangle.
For any two right-angle triangles, if one of the other two angles is equal then the third angle is also equal to each other.
If two triangles are similar to each other, then the ratio of their corresponding sides is proportional.