Question

In a quadrilateral OABC, OA=OC. The bisector of Angle AOB meets AC in D and AB in F and the bisector of angle COB meets AC in E and BC in G. Prove that ∆ODE ~ ∆OFG.

Answer 1

In any triangle, the bisector of an angle divides the opposite side in the ratio of the sides of the angle. 

∴ Considering ΔAOB, 
OA : OB = AF : BF --- (1). 

And considering ΔCOB, 
OC : OB = CG : BG --- (2). 

From (1) & (2), as OA = OC, we get, 
AF : BF = CG : BG. 

Here we can find that, F and G are the points which divide AB and BC respectively in the same ratio. 

In any triangle, the line joining two points on two sides each, which divide the sides in the same ratio, is parallel to the third side. 
∴ Considering ΔABC, 
FG || AC. 

Considering ΔODE & ΔOFG, 
∠O = ∠O (common angle) 
∠D = ∠F (corresponding angles made by the angle bisector OF which cuts the parallel lines FG & AC). 
∠E = ∠G (corresponding angles made by the angle bisector OG which cuts the parallel lines FG & AC). 

∴ ΔODE ~ ΔOFG. 

Answer 2

Answer:

In any triangle, the bisector of an angle divides the opposite side in the ratio of the sides of the angle.

∴ Considering ΔAOB,

OA : OB = AF : BF --- (1).

And considering ΔCOB,

OC : OB = CG : BG --- (2).

From (1) & (2), as OA = OC, we get,

AF : BF = CG : BG.

Here we can find that, F and G are the points which divide AB and BC respectively in the same ratio.

In any triangle, the line joining two points on two sides each, which divide the sides in the same ratio, is parallel to the third side.

∴ Considering ΔABC,

FG || AC.

Considering ΔODE & ΔOFG,

∠O = ∠O (common angle)

∠D = ∠F (corresponding angles made by the angle bisector OF which cuts the parallel lines FG & AC).

∠E = ∠G (corresponding angles made by the angle bisector OG which cuts the parallel lines FG & AC).

∴ ΔODE ~ ΔOFG.