Given that
![\begin{gathered} \angle ABC\cong\angle DEC \\ \angle GFE\cong\angle DEC \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/i4awr6modx44yzjtjopdtg0hci8lykcgor.png)
C is the point of intersection between AG and BD while E is the point of intersection between AG and DF
To prove the triangle ABC is similar to the triangle GEF
From the given figure, it is seen that
![CD=CE](https://img.qammunity.org/2023/formulas/mathematics/college/k6gkw6df6w5wi1jqr0bun6oxvawpdyps0u.png)
Therefore,
![\angle DEC=\angle DCE](https://img.qammunity.org/2023/formulas/mathematics/college/nlbtomwuwmxtax3ki18pbe1z67gnuvp5w6.png)
Since
![\begin{gathered} \angle ABC\cong\angle DEC \\ \angle GFE\cong\angle DEC \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/i4awr6modx44yzjtjopdtg0hci8lykcgor.png)
Then,
![\angle ABC\cong\angle GFE](https://img.qammunity.org/2023/formulas/mathematics/college/u87qndg801llif3tqpsmpkmuzg6fs0gz56.png)
In the triangle ABC and the triangle GFE, two angles are congruent.
Therefore,
![\begin{gathered} (AC)/(BC)=(EG)/(FE) \\ (AC)/(EG)=(BC)/(FE) \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/ce4nqmavcnn0wnkml9z1at2zq6s56peydp.png)
Similarly,
![(AB)/(FG)=(AC)/(EG)](https://img.qammunity.org/2023/formulas/mathematics/college/7dxasrcr8zwvvr05hx3w0g0we8h2ee02m3.png)
Thus,
![(AB)/(FG)=(AC)/(EG)=(BC)/(FE)](https://img.qammunity.org/2023/formulas/mathematics/college/nzefdtiilrprkllitoh7z9hsxq1d54w4hk.png)
Hence, the triangle ABC is similar to the triangle GEF