#### 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 ∆FEG respectively. If ∆ABC ~ ∆FEG then

show that

(i) \frac{CD}{GH} = \frac{AC}{FG}

(ii) ∆DCB ~ ∆HGE

Anonymous

0

show that

(i) \frac{CD}{GH} = \frac{AC}{FG}

(ii) ∆DCB ~ ∆HGE

Krishna

0

Step 1: Understand the question and According to the given information make a imaginary diagram.

Step 2: Note down the equal angles in both the triangles (ΔABC ~ ΔFEG)

NOTE: It is given that ΔABC ~ ΔFEG.

So, \angle A = \angle F ,

\angle B = \angle E

\angle B = \angle G

Step 3: Find the similar triangles hidden in the figure. (ΔACD ~ ΔFGH).

NOTE: we know that \angle ACD = \angle FGH (Angle bisector)

From this We can write \angle DCB = \angle HGE

(Angle bisector)

\angle A = \angle F (Proved above)

∴ ΔACD ~ ΔFGH (By AA similarity criterion)

Step 4: Prove the required ratio

NOTE: ΔACD ~ ΔFGH

So, their corresponding sides are in the same ratio.

\frac{CD}{GH} = \frac{AC}{FG}

Condition (i) proved.

Step 5: Find the similar figures hidden in the figure (ΔDCB ~ ΔHGE)

NOTE: In ΔDCB and ΔHGE,

\angle DCB = \angle HGE (Proved above)

\angle B = \angle E (Proved above)

∴ ΔDCB ~ ΔHGE (By AA similarity criterion)

Condition (ii) proved