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प्रश्न
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
उत्तर
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`
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