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प्रश्न
In ΔABC, PQ is a line segment intersecting AB at P and AC at Q such that seg PQ || seg BC. If PQ divides ΔABC into two equal parts having equal areas, find `"BP"/"AB"`.
उत्तर
seg PQ || seg BC and AB is their transversal.
∴∠APQ ≅ ∠ABC …(i) [Corresponding angles]
In ΔAPQ and ΔABC,
∠APQ ≅ ∠ABC …[From (i)]
∠PAQ ≅ ∠BAC …[Common angle]
∴ ΔAPQ ∼ ΔABC …[By AA test of similarity]
`("A"(Δ"APQ"))/("A"(Δ"ABC")) = "AP"^2/"AB"^2` ........(ii) [By theorem of areas of similar triangles]
A(ΔAPQ) = `1/2` A(ΔABC) ......[∵ Seg PQ divides ΔABC into two parts of equal areas.]
∴ `("A"(Δ"APQ"))/("A"(Δ"ABC")) = 1/2` .....(iii)
∴ `"AP"^2/"AB"^2 = 1/2` ...…[From (ii) and (iii)]
∴ `"AP"/"AB" = 1/sqrt2` …[Taking square root of both sides]
∴ `1 - "AP"/"AB" = 1 - 1/sqrt2` ....…[Subtracting both sides from 1]
`("AB" - "AP")/"AB" = (sqrt2 - 1)/sqrt2`
∴ `"BP"/"AB" = (sqrt2 - 1)/sqrt2` ......…[A – P – B]
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