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प्रश्न
Prove that, if a line parallel to a side of a triangle intersects the other sides in two district points, then the line divides those sides in proportion.
Prove that, 'If a line parallel to a side of a triangle intersects the remaining sides in two distinct points, then the line divides the side in the same proportion’.
उत्तर १
Given: A Δ ABC in which DE||BC, and intersect AB in D and AC in E.
To Prove: `"AD"/"BD" = "AE"/"EC"`
Construction: Join BE, CD, and draw EF⊥ BA and DG⊥CA.
Proof:
Area ∆ADE = `1/2 xx ("base" xx "height") =1/2("AD. EF")`
Area ∆DBE = `1/2 xx("base" xx "height") =1/2("DB.EF")`
⇒ `("Area"(triangle"ADE"))/("Area"(triangle"DBE"))=[((1/2"AD. EF"))/((1/2"DB. EF")]] = "AD"/"DB"`
Similarly,
⇒ `("Area"(triangle"ADE"))/("Area"(triangle"DBE"))=[((1/2"AE. DG"))/((1/2"EC. DG")]] = "AE"/"EC"`
ΔDBE and ΔDEC are on the same base DE and BC the same parallel DE and BC.
∴ Area (ΔDBE) = Area (ΔDEC)
⇒ `1/("Area"(triangle"DBE")) =1/("Area"(triangle"DEC"))` ...[Taking reciprocal of both sides]
⇒ `("Area"(triangle"ADE"))/("Area"(triangle"DBE"))=("Area"(triangle"ADE"))/("Area"(triangle"DEC")` ...[Multiplying both sides by Area (ΔADE)]
⇒ `"AD"/"DB"="AE"/"EC"`
Hence Proved.
उत्तर २
Given: In Δ ABC line l || line BC and line l intersects AB and AC in point P and Q respectively.
To prove: `"AP"/"PB" = "AQ"/"QC"`
Construction: Draw seg PC and seg BQ.
Proof: ΔAPQ and ΔPQB have equal heights.
`therefore ("A"(Delta "APQ"))/("A"(Delta "PQB")) = "AP"/"PB"` (areas proportionate to bases) ...(I)
and `("A"(Delta "APQ"))/("A"(Delta "PQC")) = "AQ"/"QC"` (areas proportionate to bases) ...(II)
Seg PQ is the common base of ΔPQB and ΔPQC, seg PQ || seg BC,
Hence ΔPQB and ΔPQC have equal areas.
A(ΔPQB) = A(ΔPQC) ...(III)
`("A"(Delta "APQ"))/("A"(Delta "PQB")) = ("A"(Delta "APQ"))/("A"(Delta "PQC"))` ...[from (I), (II) and (III)]
`therefore "AP"/"PB" = "AQ"/"QC"` ...[from (I) and (II)]
Hence Proved.
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In the given figure, in ∆ABC, point D on side BC is such that, ∠BAC = ∠ADC. Prove that, CA2 = CB × CD
A line is parallel to one side of triangle which intersects remaining two sides in two distinct points then that line divides sides in same proportion.
Given: In ΔABC line l || side BC and line l intersect side AB in P and side AC in Q.
To prove: `"AP"/"PB" = "AQ"/"QC"`
Construction: Draw CP and BQ
Proof: ΔAPQ and ΔPQB have equal height.
`("A"(Δ"APQ"))/("A"(Δ"PQB")) = (["______"])/"PB"` .....(i)[areas in proportion of base]
`("A"(Δ"APQ"))/("A"(Δ"PQC")) = (["______"])/"QC"` .......(ii)[areas in proportion of base]
ΔPQC and ΔPQB have [______] is common base.
Seg PQ || Seg BC, hence height of ΔAPQ and ΔPQB.
A(ΔPQC) = A(Δ______) ......(iii)
`("A"(Δ"APQ"))/("A"(Δ"PQB")) = ("A"(Δ "______"))/("A"(Δ "______"))` ......[(i), (ii), and (iii)]
`"AP"/"PB" = "AQ"/"QC"` .......[(i) and (ii)]
In the given figure ΔABC ~ ΔPQR, PM is median of ΔPQR. If ar ΔABC = 289 cm², BC = 17 cm, MR = 6.5 cm then the area of ΔPQM is ______.
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Construct an equilateral triangle of side 7 cm. Now, construct another triangle similar to the first triangle such that each of its sides are `5/7` times of the corresponding sides of the first triangle.
Prove that If a line is drawn parallel to one side of a triangle to intersect the other two sides in distinct points, the other two sides are divided in the same ratio. In the figure, find EC if `(AD)/(DB) = (AE)/(EC)` using the above theorem.
State and prove Basic Proportionality theorem.
