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प्रश्न
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.
उत्तर
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`
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