In the picture below, the side AC of the triangle ABC is extended to D

1.	In the picture below, the side AC of the triangle ABC is extended to D, by adding the length of the side CB. Then the line through C parallel to DB is drawn to meet AB at E.1. Prove that CE bisects ∠C.2. Describe how this can be used to divide an 8 centimeters long line in the ratio 4 : 5.3. Can we use it to divide an 8 centimeters long line in the ratio 3 : 4? How?
Answer» 1. ∆ BCD is an equilateral triangle If ∠CBD = x then ∠CDB = x ∠BCD = 180 – 2x ∠BCE = x (line BC passing through parallel lines EC and BD makes equal angle) ∴ ∠ACE = x ∴ line CE bisects ∠C. 2. If AB = 8, AC = 4, and CB = 5 then the line CE divides AB in the ratio 4 : 5. 3. No. We can’t draw a triangle with sides 8cm, 3cm and 4cm. Since 3 : 4 = 6 : 8, take 6cm and 8cm as two sides then find third side.

In the picture below, the side AC of the triangle ABC is extended to D, by adding the length of the side CB. Then the line through C parallel to DB is drawn to meet AB at E.1. Prove that CE bisects ∠C.2. Describe how this can be used to divide an 8 centimeters long line in the ratio 4 : 5.3. Can we use it to divide an 8 centimeters long line in the ratio 3 : 4? How?

Answer»

1. ∆ BCD is an equilateral triangle If ∠CBD = x then ∠CDB = x

∠BCD = 180 – 2x

∠BCE = x

(line BC passing through parallel lines EC and BD makes equal angle)

∴ ∠ACE = x

∴ line CE bisects ∠C.

2. If AB = 8, AC = 4, and CB = 5 then the line CE divides AB in the ratio 4 : 5.

3. No.

We can’t draw a triangle with sides 8cm, 3cm and 4cm. Since 3 : 4 = 6 : 8, take 6cm and 8cm as two sides then find third side.