Question

D is the mid-point of side BC of a ΔABC. AD is bisected at the point E and BE produced cuts AC at the point X. Prove that BE : EX = 3 : 1

Answer 1

Given: In ΔABC, D is the mid-point of BC and E is the mid-point of AD.

To prove: BE : EX = 3 : 1

Const: Through D, draw DF || BX

Proof: In ΔEAX and ΔADF

∠EAX = ∠ADF                         [Common]

∠AXE = ∠DAF                         [Corresponding angles]

Then, ΔAEX ~ ΔADF                [By AA similarity]

`therefore"EX"/"DF"="AE"/"AD"`                      [Corresponding parts of similar Δ are proportional]

`rArr"EX"/"DF"="AE"/"2AE"`           [AE = ED given]

⇒ DF = 2EX …. (i)

In ΔCDF and ΔCBX               [By AA similarity]

`therefore"CD"/"CB"="DF"/"BX"`                [Corresponding parts of similar Δ are proportional]

`rArr1/2="DF"/"BE + EX"`         [BD = DC given]

⇒ BE + EX = 2DF

⇒ BE + EX = 4EX

⇒ BE = 4EX – EX                         [By using (i)]

⇒ BE = 4EX – EX

`rArr"BE"/"EX"=3/1`