Question

In ∆ABC, E is the mid-point of the median AD, and BE produced meets side AC at point Q.

Show that BE: EQ = 3: 1.

Answer 1

Construction: Draw DX || BQ

In ΔBCQ and ΔDCX,

∠BCQ = ∠DCX                ...(Common)

∠BQC = ∠DXC                ...(Corresponding angles)

So, ΔBCQ ∼ ΔDCX          ....(AA Similarity criterion)

⇒ `"BQ"/"DX" = "BC"/"DC" = "CQ"/"CX"`       ...(Corresponding sides are proportional.)

⇒ `"BQ"/"DX" = "2CD"/"CD"`         ...(D is the mid-point of BC)    

⇒ `"BQ"/"DX" = 2`                 ...(i)

Similarly, ΔAEQ ∼ ΔADX,

⇒ `"EQ"/"DX" = "AE"/"ED" = 1/2`      ...(E is the mid-point of AD)

That is `"EQ"/"DX" = 1/2`              ...(ii)

Dividing (i) by (ii), We get

⇒ `"BQ"/"EQ" = 4`

⇒ BE + EQ = 4EQ

⇒ BE = 3EQ

⇒ `"BQ"/"EQ" = 3/1`