মনে করা যাক, \(\frac{a}{b} = \frac{c}{d} = k(k \ne 0)\)
\(\therefore a=b k, c=d k\) (এখানে k একটি অশূন্য ধ্রুবক)
এখন, \(\frac{{{a^2} + {b^2}}}{{{a^2} - {b^2}}} = \frac{{{{(bk)}^2} + {b^2}}}{{{{(bk)}^2} - {b^2}}} = \frac{{{b^2}\left( {{k^2} + 1} \right)}}{{{b^2}\left( {{k^2} - 1} \right)}} = \frac{{{k^2} + 1}}{{{k^2} - 1}}(b \ne 0)\)
আবার, \(\frac{{ac + bd}}{{ac - bd}} = \frac{{bk \times dk + bd}}{{bk\times dk-bd}} = \frac{{bd\left( {{k^2} + 1} \right)}}{{bd\left( {{k^2} - 1} \right)}} = \frac{{{k^2} + 1}}{{{k^2} - 1}}{\rm{ }}(bd \ne 0)\)
সুতরাং হল যে, \(\left( {{a^2} + {b^2}} \right):\left( {{a^2} - {b^2}} \right) = (ac + bd):(ac - bd)\) (প্রমাণিত)

\(\frac{{{a^2} + ab + {b^2}}}{{{a^2} - ab + {b^2}}} = \frac{{{{(bk)}^2} + bk \times b + {b^2}}}{{{{(bk)}^2} - bk \times b + {b^2}}} = \frac{{{b^2}\left( {{k^2} + k + 1} \right)}}{{{b^2}\left( {{k^2} - k + 1} \right)}} = \frac{{{k^2} + k + 1}}{{{k^2} - k + 1}}\)
আবার, \(\frac{{{c^2} + cd + {d^2}}}{{{c^2} - cd + {d^2}}} = \frac{{{{(dk)}^2} + dk \times d + {d^2}}}{{{{(dk)}^2} - dk \times d + {d^2}}} = \frac{{{d^2}\left( {{k^2} + k + 1} \right)}}{{{d^2}\left( {{k^2} - k + 1} \right)}} = \frac{{{k^2} + k + 1}}{{{k^2} - k + 1}}\)
তাহলে প্রমাণিত হল যে, \(\left( {{a^2} + ab + {b^2}} \right):\left( {{a^2} - ab + {b^2}} \right) = \left( {{c^2} + cd + {d^2}} \right):\left( {{c^2} - cd + {d^2}} \right)\) (প্রমাণিত)

\(\frac{{\sqrt {{a^2} + {c^2}} }}{{\sqrt {{b^2} + {d^2}} }} = \frac{{\sqrt {{(bk)}^2 + {{(dk)}^2}}}}{{\sqrt {{b^2} + {d^2}} }} = \frac{{\sqrt {{b^2}{k^2} + {d^2}{k^2}} }}{{\sqrt {{b^2} + {d^2}} }} = \frac{{k\sqrt {{b^2} + {d^2}} }}{{\sqrt {{b^2} + {d^2}} }} = k\)
আবার, \(\frac{{pa + qc}}{{pb + qd}} = \frac{{p.bk + qdk}}{{pb + qd}} = \frac{{k(pb + qd)}}{{pb + qd}} = k\)
তাহলে প্রমাণিত হল যে, \(\sqrt {{a^2} + {c^2}} :\sqrt {{b^2} + {d^2}} = (pa + qc):(pb + qd)\) (প্রমাণিত)

\(x: a=y: b=z: c\)
মনে করা যাক, \(\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k(k \ne 0)\)
তাহলে \(x = ak,y = bk,z = ck\)
বামপক্ষ \( = \frac{{{x^3}}}{{{a^2}}} + \frac{{{y^3}}}{{{b^2}}} + \frac{{{z^3}}}{{{c^2}}} = \frac{{{{(ak)}^3}}}{{{a^2}}} + \frac{{{{(bk)}^3}}}{{{b^2}}} + \frac{{{{(ck)}^3}}}{{{c^2}}} = \frac{{{a^3}{k^3}}}{{{a^2}}} + \frac{{{b^3}{k^3}}}{{{b^2}}} + \frac{{{c^3}{k^3}}}{{{c^2}}}\)
\( = {a{k^3}} + b{k^3} + c{k^3} = {k^3}(a + b + c)\)
ডানপক্ষ \( = \frac{{{{(x + y + z)}^3}}}{{{{(a + b + c)}^2}}} = \frac{{{{(ak + bk + ck)}^3}}}{{{{(a + b + c)}^2}}} = \frac{{{k^3}{{(a + b + c)}^3}}}{{{{(a + b + c)}^2}}} = {k^3}(a + b + c)\)
সুতরাং, বামপক্ষ = ডানপক্ষ (প্রমাণিত)।

বামপক্ষ \(\frac{{{x^3} + {y^3} + {z^3}}}{{{a^3} + {b^3} + {c^3}}} = \frac{{{{(ak)}^3} + {{(bk)}^3}{{(ck)}^3}}}{{{a^3} + {b^3} + {c^3}}} = \frac{{{k^3}\left( {{a^3} + {b^3} + {c^3}} \right)}}{{{a^3} + {b^3} + {c^3}}} = {k^3}\)
ডানপক্ষ \(\frac{{xyz}}{{abc}} = \frac{{ak \cdot bkck}}{{abc}} = \frac{{{k^3}abc}}{{abc}} = {k^3}\)
সুতরাং বামপক্ষ = ডানপক্ষ (প্রমাণিত)।

বামপক্ষ \( = \left( {{a^2} + {b^2} + {c^2}} \right)\left( {{x^2} + {y^2} + {z^2}} \right)\)
\(=\left(a^2+b^2+c^2\right)\left\{(a k)^2+(b k)^2+(c k)^2\right\}\)
\( = \left( {{a^2} + {b^2} + {c^2}} \right)\left( {{a^2}{k^2} + {b^2}{k^2} + {c^2}{k^2}} \right)\)
\( = \left( {{a^2} + {b^2} + {c^2}} \right) \times {k^2} \times \left( {{a^2} + {b^2} + {c^2}} \right)\)
\( = {k^2}{\left( {{a^2} + {b^2} + {c^2}} \right)^2}\)
ডানপক্ষ \( = {(ax + by + cz)^2} = {(a \times ak + b \times bk + c \times ck)^2}\)
\( = {\left\{ {k\left( {{a^2} + {b^2} + {c^2}} \right)} \right\}^2} = {k^2}{\left( {{a^2} + {b^2} + {c^2} } \right)^2}\)
সুতরাং বামপক্ষ = ডানপক্ষ (প্রমাণিত)।

\(a: b=c: d=e: f\) [এখানে k একটি অশূন্য ধ্রুবক]
মনে করি,
\(\frac{a}{b}=\frac{c}{d}=\frac{e}{f}=k \)
\(\therefore a=b k, c=d k, e=f k\)
এখন, \(\frac{5 a-7 c-13 e}{5 b-7 d-13 f}=\frac{5 b k-7 d k-13 f k}{5 b-7 d-13 f}\)
\(=\frac{k(5 b-7 d-13 f)}{5 b-7 d-13 f}=k=\) প্রত্যেকটি অনুপাত।(প্রমাণিত)

বামপক্ষ\( = \left( {{a^2} + {c^2} + {e^2}} \right)\left( {{b^2} + {d^2} + {f^2}} \right)\)
\( = \left\{ {{{(bk)}^2} + {{(dk)}^2} + {{(fk)}^2}} \right\}\left( {{b^2} + {d^2} + {f^2}} \right)\)
\( = \left( {{b^2}{k^2} + {d^2}{k^2} + {f^2}{k^2}} \right)\left( {{b^2} + {d^2} + {f^2}} \right)\) \([\because a=b k, c=d k, e=f k]\)
\( = {{\rm{k}}^2}\left( {{b^2} + {d^2} + {f^2}} \right)\left( {{b^2} + {d^2} + {f^2}} \right) \)
\(= {k^2}{\left( {{b^2} + {d^2} + {f^2}} \right)^2}\)
ডানপক্ষ\( = {(ab + cd + {\rm{ ef }})^2}\)\( = {(bk \times b + dk \times d + fk \times f)^2}\)
\( = {\left\{ {k\left( {{b^2} + {d^2} + {f^2}} \right)} \right\}^2} = {k^2}{\left( {{b^2} + {d^2} + {f^2}} \right)^2}\)
\(\therefore\) বামপক্ষ = ডানপক্ষ (প্রমাণিত) ।

মনে করা যাক, \(\frac{a}{b} = \frac{b}{c} = k(k \ne 0)\)
\(\therefore b = ck\)এবং \(a = bk = ck.k = c{k^2}\)
বামপক্ষ \({\left( {\frac{{a + b}}{{b + c}}} \right)^2} = {\left( {\frac{{c{k^2} + ck}}{{ck + c}}} \right)^2} = {\left\{ {\frac{{ck(k + 1)}}{{c(k + 1)}}} \right\}^2} = \frac{{{c^2}{k^2}}}{{{c^2}}} = {k^2}\)
ডানপক্ষ \(\frac{{{a^2} + {b^2}}}{{{b^2} + {c^2}}} = \frac{{{{\left( {c{k^2}} \right)}^2} + {{(ck)}^2}}}{{{{(ck)}^2} + {c^2}}} = \frac{{{c^2}{k^4} + {c^2}{k^2}}}{{{c^2}{k^2} + {c^2}}} = \frac{{{c^2}{k^2}\left( {{k^2} + 1} \right)}}{{{c^2}\left( {{k^2} + 1} \right)}} = {k^2}\)
সুতরাং, বামপক্ষ = ডানপক্ষ (প্রমাণিত)।

যেহেতু a, b, c ক্রমিক সমানুপাতী, অতএব
\(\frac{a}{b}=\frac{b}{c}=k(\neq 0)\) (ধরি)
\(\therefore a=b k, \quad b=c k\)
অর্থাৎ \(a=c k . k=c k^{2}\) এবং \(b = ck\)
\(\therefore\) বামপক্ষ \(=a^{2} b^{2} c^{2}\left(\frac{1}{a^{3}}+\frac{1}{b^{3}}+\frac{1}{c^{3}}\right)\) \(=a^2 b^2 c^2\left(\frac{1}{\left(c k^2\right)^3}+\frac{1}{(c k)^3}+\frac{1}{c^3}\right)\)
\(=c^{2} k^{4} \times c^{2} k^{2} \times c^{2}\left(\frac{1}{c^{3} k^{6}}+\frac{1}{c^{3} k^{3}}+\frac{1}{c^{3}}\right)\)
\(=c^{6} k^{6} \times \frac{1}{c^{3}}\left(\frac{1}{k^{6}}+\frac{1}{k^{3}}+1\right)\)
\(=c^{3} k^{6} \times \frac{1+k^{3}+k^{6}}{k^{6}}\)
\(=c^{3}\left(1+k^{3}+k^{6}\right)\)
\(=c^{3}+c^{3} k^{3}+c^{3} k^{6}\)
\(=\left(c k^{2}\right)^{3}+(c k)^{3}+c^{3}\)
\(=a^{3}+b^{3}+c^{3}\)
\(\left(\because a=c k^{2}, b=c k\right)\)
\(\therefore\) বামপক্ষ = ডানপক্ষ (প্রমাণিত)

বামপক্ষ\( = \frac{{abc{{(a + b + c)}^3}}}{{{{(ab + bc + ca)}^3}}} \)
\(\left[\begin{array}{rl}\because a & =c k^2 \\ b & =c k\end{array}\right]\)
\(= \frac{{c{k^2} \times ck \times c{{\left( {c{k^2} + ck + c} \right)}^3}}}{{{{\left( {c{k^2} \times ck + ck \times c + c \times c{k^2}} \right)}^3}}}\)
\( = \frac{{{c^3}{k^3}{{\left\{ {c\left( {{k^2} + k + 1} \right)} \right\}}^3}}}{{{{\left( {{c^2}{k^3} + {c^2}k + {c^2}{k^2}} \right)}^3}}} = \frac{{{c^6}{k^3}{{\left( {{k^2} + k + 1} \right)}^3}}}{{{{\left\{ {{c^2}k\left( {{k^2} + 1 + k} \right)} \right\}}^3}}}\)
\( = \frac{{{c^6}{k^3}{{\left( {{k^2} + k + 1} \right)}^3}}}{{{c^6}{k^3}{{\left( {{k^2} + k + 1} \right)}^3}}} = 1\)(প্রমাণিত)।

\(a, b, c, d\)
যেহেতু, a, b, c, d ক্ৰমিক সমানুপাতী,
\(\frac{a}{b} = \frac{b}{c} = \frac{c}{d} = k\)মনে করা যাক
\(\therefore c = dk, b = ck = dk \times k =dk^2,a = bk = dk^2 \times k ={k}^3\)
বামপক্ষ\( = \left( {{a^2} + {b^2} + {c^2}} \right)\left( {{b^2} + {c^2} + {d^2}} \right)\)
\( = \left\{ {{{\left( {d{k^3}} \right)}^2} + {{\left( {d{k^2}} \right)}^2} + {{(dk)}^2}} \right\}\left\{ {{{\left( {d{k^2}} \right)}^2} + {{(dk)}^2} + {d^2}} \right\}\)
\( = \left\{ {{{\rm{d}}^2}{{\rm{k}}^6} + {{\rm{d}}^2}{{\rm{k}}^4} + {{\rm{d}}^2}{{\rm{k}}^2}} \right\}\left\{ {{{\rm{d}}^2}{{\rm{k}}^4} + {{\rm{d}}^2}{{\rm{k}}^2} + {{\rm{d}}^2}} \right\}\)
\( = {{\rm{d}}^2}{{\rm{k}}^2}\left( {{{\rm{k}}^4} + {{\rm{k}}^2} + 1} \right).{{\rm{d}}^2}\left( {{{\rm{k}}^4} + {{\rm{k}}^2} + 1} \right)\)
\(= {{\rm{d}}^4}{{\rm{k}}^2}{\left( {{{\rm{k}}^4} + {{\rm{k}}^2} + 1} \right)^2}\)
ডানপক্ষ \( = {(ab + bc + cd)^2}\)
\(= {\left\{ {d{k^3} \times d{k^2} + d{k^2} \times dk + dk \times d} \right\}^2}\)
\( = {\left( {{d^2}{k^5} + {d^2}{k^3} + {d^2}k} \right)^2}\)
\(= {\left\{ {{d^2}k\left( {{k^4} + {k^2} + 1} \right)} \right\}^2}\)
\( = {d^4}{k^2}{\left( {{k^4} + {k^2} + 1} \right)^2}\)
\(\therefore\) বামপক্ষ = ডানপক্ষ (প্রমাণিত)।

বামপক্ষ \( = {(b - c)^2} + {(c - a)^2} + {(b - d)^2}\)
\( = {\left( {d{k^2} - dk} \right)^2} + {\left( {dk - d{k^3}} \right)^2} + {\left( {d{k^2} - d} \right)^2}\) \(\left[\because a=d k^3, b=d k^2, c=d k\right]\)
\( = {\{ dk(k - 1)\} ^2} + \{ dk{\left( {1 - {k^2}} \right)\}^2} + {\left\{ {d\left( {{k^2} - 1} \right)} \right\}^2}\)
\( = {{\rm{d}}^2}{k^2}\left( {{k^2} - 2k + 1} \right) + {d^2}{k^2}\left( {1 - 2{k^2} + {k^4}} \right) + {d^2}\left( {{k^4} - 2{k^2} + 1} \right)\)
\( = {{\rm{d}}^2}{k^2}\left( {{k^2} - 2k + 1} \right) + {d^2}{k^2}\left( {1 - 2{k^2} + {k^4}} \right) + {d^2}\left( {{k^4} - 2{k^2} + 1} \right)\)
\( = {{\rm{d}}^2}\left\{ {{{\rm{k}}^4} - \left( {2{{\rm{k}}^3}} \right) + {{\rm{k}}^2} + {{\rm{k}}^2} - 2{{\rm{k}}^4} + \left( {{{\rm{k}}^6}} \right) + {{\rm{k}}^4} - 2{{\rm{k}}^2} + 1} \right\}\)
\( = {d^2}\left( {{k^6} - 2{k^3} + 1} \right)\)
\(=d^2\left\{\left(k^3\right)^2-2 \times k^3 \times 1+(1)^2\right\}\)
\(= {d^2}{\left( {{k^3} - 1} \right)^2}\)
ডানপক্ষ\( = {(a - d)^2}\)
\(= {\left( {d{k^3} - d} \right)^2}\)
\(= {\left\{ {d\left( {{k^3} - 1} \right)} \right\}^2}\)
\(= {d^2}{\left( {{k^3} - 1} \right)^2}\)
\(\therefore\) বামপক্ষ = ডানপক্ষ (প্রমাণিত)।

\(\frac{m}{a} = \frac{n}{b} = k(k \ne 0)\) ধরা যাক তাহলে, \({\rm{m}} = {\rm{ak}},{\rm{n}} = {\rm{bk}}\)
বামপক্ষ\( = \left( {{m^2} + {n^2}} \right)\left( {{a^2} + {b^2}} \right)\)
\(= \left\{ {{{(ak)}^2} + {{(bk)}^2}} \right\}\left( {{a^2} + {b^2}} \right)\) \([\because m=a k, n=b k]\)
\( = \left( {{a^2}{k^2} + {b^2}{k^2}} \right)\left( {{a^2} + {b^2}} \right)\)
\( = {k^2}\left( {{a^2} + {b^2}} \right)\left( {{a^2} + {b^2}} \right)\)
\( = {k^2}{\left( {{a^2} + {b^2}} \right)^2}\)
ডানপক্ষ\( = {(am + bn)^2}\)
\( = {(a \times ak + b \times bk)^2}\)
\( = {\left\{ {k\left( {{a^2} + {b^2}} \right)} \right\}^2}\)
\( = {{\rm{k}}^2}{\left( {{{\rm{a}}^2} + {{\rm{b}}^2}} \right)^2}\)
\(\therefore\) বামপক্ষ = ডানপক্ষ (প্রমাণিত)
অর্থাৎ, \(\left( {{m^2} + {n^2}} \right)\left( {{a^2} + {b^2}} \right) = {(am + bn)^2}\) (প্রমাণিত)।

এখানে দেওয়া আছে \(\frac{a}{b} = \frac{x}{y}\)
বা, \(\frac{a}{x} = \frac{b}{y} = k( \ne 0)\) হলে ধরা যাক
\(\therefore a = kx,b = ky\)
বামপক্ষ \(=(a y+b)\left(a^2+b^2\right) x^3\)
\(=(k x+k y)\left\{k^2\left(x^2+y^2\right)\right\} \times x^3\)
\( = k(x + y)\left\{ {{k^2}\left( {{x^2} + {y^2}} \right)} \right\} \cdot {x^3}\)
\( = (a + b)\left( {{a^2} + {b^2}} \right){x^3}\)
\(= (kx + ky)\left\{ {{{(kx)}^2} + {{(ky)}^2}} \right\}{x^3}\)
\(=k(x+y) \{k^{2} \left( x^{2}+y^{2}\right)\} \cdot x^{3}\)
\( = {k^3}(x + y)\left( {{x^2} + {y^2}} \right){x^3}\)
\( = (x + y)\left( {{x^2} + {y^2}} \right){(kx)^3}\)
\( = (x + y)\left( {{x^2} + {y^2}} \right){(a)^3}\)
\(= (x + y)\left( {{x^2} + {y^2}} \right){a^3}\)
= ডানপক্ষ (প্রমাণিত)।
\(\therefore\) বামপক্ষ = ডানপক্ষ (প্রমাণিত)।

ধরি, \(\frac{x}{l m-n^{2}}=\frac{y}{m n-l^{2}}=\frac{z}{n l-m^{2}}=k\)
\(\therefore \quad x=k\left(l m-n^{2}\right), y=k\left(m n-l^{2}\right), z=k\left(n l-m^{2}\right)\)
এখন, \(l x+m y+n z\)
\(=l k\left(l m-n^{2}\right)+m k\left(m n-l^{2}\right)+n k\left(n l-m^{2}\right)\)
\(=k\left\{l\left(l m-n^{2}\right)+m\left(m n-l^{2}\right)+n\left(n l-m^{2}\right)\right\}\)
\(=k\left(l^{2} m-ln ^{2}+m^{2} n-l^{2} m+n^{2} l-m^{2} n\right)\)
\(=k \times 0=0\)
অতএব, \(l x+m y+n z=0\) (প্রমাণিত)

মনে করা যাক, \(\frac{x}{{b + c - a}} = \frac{y}{{c + a - b}} = \frac{z}{{a + b - c}} = k( \ne 0)\)
\(\therefore x = k(b + c - a),y = k(c + a - b),z = k(a + b - c)\)
এখন, \({\rm{(b - c ) }}x{\rm{ + ( c - a ) y + ( a - b ) z}}\)
\({\rm{ = ( b - c ) }}{\rm{ k ( b + c - a ) + ( c - a ) k ( c + a - b ) + ( a - b ) k ( a + b - c )}}\)
\( = k\{ (b - c)(b + c - a) + (c - a)(c + a - b) + (a - b)(a + b - c)\} \)
\( = k \{ (b - c)(b + c) - a(b - c) + (c - a)(c + a) - b(c - a) + (a - b)(a + b) - c(a - b)\}\)
\( = k\left\{ {{b^2} - {c^2} - ab + ca + {c^2} - {a^2} - bc + ab + {a^2} - {b^2} - ca + bc} \right\} = k \times 0\)
\(= 0\)(প্রমাণিত)

দেওয়া আছে, \(\frac{x}{y} = \frac{{a + 2}}{{a - 2}}\) দুদিকের বর্গ করলে, \(\frac{{{x^2}}}{{{y^2}}} = \frac{{{{(a + 2)}^2}}}{{{{(a - 2)}^2}}}\)
বা, \(\frac{{{x^2} + {y^2}}}{{{x^2} - {y^2}}} = \frac{{{{(a + 2)}^2} + {{(a - 2)}^2}}}{{{{(a + 2)}^2} - {{(a - 2)}^2}}}\) [উভয়দিকে যোগ-ভাগ করে পাই]
\( = \frac{{{a^2} + 4a + 4 + {a^2} - 4a + 4}}{{{a^2} + 4a + 4 - {a^2} + 4a - 4}}\)
\( = \frac{{2\left( {{a^2} + 4} \right)}}{{8a}}\)
\(= \frac{{{a^2} + 4}}{{4a}}\)
\(\therefore\) \(\frac{{{x^2} - {y^2}}}{{{x^2} + {y^2}}} = \frac{{4a}}{{{a^2} + 4}}\) (প্রমাণিত)

\(x = \frac{{8ab}}{{a + b}}\)
বা, \(\frac{x}{{4a}} = \frac{{2b}}{{a + b}}\) (উভয় পক্ষকে 4a দ্বারা ভাগ করে পাই)
সুতরাং \(\frac{{x + 4a}}{{x - 4a}} = \frac{{2b + a + b}}{{2b - a - b}}\) [যোগ-ভাগ প্রক্রিয়ার সাহায্যে পাই]
\( = \frac{{3b + a}}{{b - a}}\)
আবার, \(x = \frac{{8ab}}{{a + b}}\)
বা, \(\frac{x}{{4b}} = \frac{{2a}}{{a + b}}\) (উভয়পক্ষকে 4b দ্বারা ভাগ করে পাই)
সুতরাং \(\frac{{x + 4b}}{{x - 4b}} = \frac{{2a + a + b}}{{2a - a - b}} = \frac{{3a + b}}{{a - b}}\) [জগ-ভাগ প্রক্রিয়ার সাহায্যে পাই]
\(\therefore\) \(\frac{{x + 4a}}{{x - 4a}} + \frac{{x + 4b}}{{x - 4b}}\) \(= \frac{{3b + a}}{{b - a}} + \frac{{3a + b}}{{a - b}}\)
\(=\frac{3 b+a}{b-a}+\frac{3 a+b}{-(b-a)}\)
\(= \frac{{3{b} + a}}{{b - a}} - \frac{{3a + b}}{{b - a}}\)
\( = \frac{{3b + a - 3a - b}}{{b - a}}\)
\(= \frac{{2b - 2a}}{{b - a}}\)
\(= \frac{{2(b - a)}}{{(b - a)}}\)
\(\therefore\) নির্ণেয় মান \(= 2\)

মনে করি, \(\frac{a}{3}=\frac{b}{4}=\frac{c}{7}=k(\neq 0)\)
\(\therefore a=3 k, \quad b=4 k, \quad c=7 k\)
সুতরাং, \(\frac{a+b+c}{c}=\frac{3 k+4 k+7 k}{7 k}=\frac{14 k}{7 k}=2\)
\(\therefore\) নির্ণেয় মান \(= 2\) (প্রমাণিত)

মনে করা যাক, \(\frac{a}{q-r}=\frac{b}{r-p}=\frac{c}{p-q}=k\) \((\because k \neq 0)\)
তাহলে, \({\rm{a = k ( q - r ) , b = k ( r - p ) , c = k ( p - q )}}\)
এখন, \({\rm{a + b + c = k ( q - r ) + k ( r - p ) + k ( p - q )}}\)
\( = {\rm{k}}({\rm{q}} - {\rm{r}} + {\rm{r}} - {\rm{p}} + {\rm{p}} - {\rm{q}}) = {\rm{k}} \times 0 = 0\)
আবার, \({\rm{pa}} + {\rm{qb}} + {\rm{rc}} = {\rm{p}} \times {\rm{k}}({\rm{q}} - {\rm{r}}) + q \times {\rm{k}}({\rm{r}} - {\rm{p}}) + {\rm{r}} \times k({\rm{p}} - {\rm{q}})\)
\( = {\rm{k}}\{ {\rm{p}}({\rm{q}} - {\rm{r}}) + {\rm{q}}({\rm{r}} - {\rm{p}}) + {\rm{r}}({\rm{p}} - q)\}\)
\( = k(pq - rp + qr - pq + rp - qr) = k \times 0 = 0\)
তাহলে দেখানো হল যে, \({\rm{a + b + c = 0 = p a + q b + r c}}\) (প্রমাণিত)

প্রদত্ত \(\frac{a x+b y}{a}=\frac{b x-a y}{b}\)
বা, \(\frac{a(a x+b y)}{a \cdot a}=\frac{b(b x-a y)}{b b}\)
বা, \(\frac{a^{2} x+a b y}{a^{2}}=\frac{b^{2} x-a b y}{b^{2}}=\frac{a^{2} x+a b y+b^{2} x-a b y}{a^{2}+b^{2}}\) [সংযােজন প্রক্রিয়া।]
\(=\frac{\left(a^{2}+b^{2}\right) x}{a^{2}+b^{2}}=x\)
\(\therefore \frac{a x+b y}{a}=\frac{b x-a y}{b}=x\)(প্রমাণিত)
বিকল্প পদ্ধতি :
মনে করি, \(\frac{a x+b y}{a}=\frac{b x-a y}{b}=k\)\([\because k \neq 0]\)
\(\therefore k a=a x+b y, k b=b x-a y\)
\(\therefore k a \cdot a+k b b=a(a x+b y)+b(b x-a y)\)
বা, \(k\left(a^{2}+b^{2}\right)=a^{2} x+a b y+b^{2} x-a b y\)
\(=a^{2} x+b^{2} x=x\left(a^{2}+b^{2}\right)\)
\(\therefore k=\frac{x\left(a^{2}+b^{2}\right)}{a^{2}+b^{2}}=x\)
\(\therefore\) প্রতিটি অনুপাত = \(k = x\) (প্রমাণিত)

একান্তর প্রক্রিয়ানুসারে, \(\frac{a+b}{c+d}=\frac{b+c}{d+a}\)
বা, উভয়পক্ষে 1 যােগ করে পাই, \(\frac{a+b}{c+d}+1=\frac{b+c}{d+a}+1\)
বা, \(\frac{a+b+c+d}{c+d}=\frac{a+b+c+d}{d+a}\)
বা, \((a+b+c+d)\left(\frac{1}{c+d}-\frac{1}{d+a}\right)=0\)
\(\therefore\) হয় \(a + b + c + d = 0 \cdots(i)\)
বা, \(\frac{1}{c+d}=\frac{1}{d+a}\)
অর্থাৎ, \(c+ d = d + a \cdots(ii)\)
অর্থাৎ, c = a
অতএব, (i) এবং (ii) থেকে
হয় \(c = a\) কিংবা \(a = b + c+ d = 0\) (প্রমাণিত)

মনে করি, \(\frac{x}{b+c}=\frac{y}{c+a}=\frac{2}{a+b}=k\) [যেখানে k একটি সমানুপাতিক ধ্রুবক]
\(\therefore x=k(b+c), y=k(c+a), z=k(a+b)\)
এখন, \(\frac{a}{y+z-x}\)
\(=\frac{a}{k(c+a)+k(a+b)-k(b+c)}\)
\(=\frac{a}{k(c+a+a+b-b-c)}=\frac{a}{k \cdot 2 a}=\frac{1}{2 k}\)
আবার, \(\frac{b}{z+x-y}\) \(=\frac{b}{k(a+b)+k(b+c)-k(c+a)}\)
\(=\frac{b}{k(a+b+b+c-c-a)}=\frac{b}{k \cdot 2 b}=\frac{1}{2 k}\)
\(\frac{c}{x+y-z}\)
আবার, \(=\frac{c}{k(b+c)+k(c+a)-k(a+b)}\)
\(=\frac{c}{k(b+c+c+a-a-b)}=\frac{c}{k \cdot 2 c}=\frac{1}{2 k}\)
\(\therefore \frac{a}{y+z-x}=\frac{b}{z+x-y}=\frac{c}{x+y-z}=\frac{1}{2 k}\)
অতএব, \(\frac{a}{x+z-x}=\frac{b}{z+x-y}=\frac{c}{x+y-z}\) (প্রমাণিত)

মনে করি, \(\frac{x+y}{3 a-b}=\frac{y+z}{3 b-c}=\frac{z+x}{3 c-a}=k\) [যেখানে k একটি সমানুপাতিক ধ্রুবক]
\(\therefore x+y=k(3 a-b), y+z=k(3 b-c), z+x=k(3 c-a)\)
\(\therefore(x+y)+(y+z)+(z+x)=k\{(3 a-b)+(3 b-c)+(3 c-a)\}\)
বা, \(2(x+y+z) = 2k(a+b+c)\)
বা, \(x+y+z = k(a+b+c)\).....(1)
\(\therefore x=k(a+b+c)-k(3 b-c)=k(a-2 b+2 c)\)
\(y=k(a+b+c)-k(3 c-a)=k(2 a+b-2 c)\)
\(z=k(a+b+c)-k(3 a-b)=k(2 b-2 a+c)\)
\(\therefore \frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}\)
\(=\frac{a\{k(a-2 b+2 c)\}+b\{k(2 a+b-2 c)\}+c\{k(2 b-2 a+c)\}}{a^2+b^2+c^2}\)
\(=\frac{k\left(a^{2}-2 a b+2 a c+2 a b+b^{2}-2 b c+2 b c-2 a c+c^{2}\right)}{a^{2}+b^{2}+c^{2}}\)
\(=\frac{k\left(a^2+b^2+c^2\right)}{a^2+b^2+c^2}\)
\(=\frac{k\left(a^{2}+b^{2}+c^{2}\right)}{a^{2}+b^{2}+c^{2}}=k \ldots \ldots\) (2)
(1) ও (2) থেকে পাই \(\frac{x+y+z}{a+b+c}=k=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}\)
অর্থাৎ, \(\frac{x+y+z}{a+b+c}=\frac{a x+b y+c z}{a^{2}+b^{2}+c^{2}}\) (প্রমাণিত)\(\therefore\)

মনে করা যাক, \(\frac{x}{a} = \frac{y}{b} = \frac{z}{c} = k( \ne 0)\) [যেখানে k একটি সমানুপাতিক ধ্রুবক]
\(\therefore x=a k, y=b k, z=c k\)
এখন, \(\frac{{{x^2} - yz}}{{{a^2} - bc}} = \frac{{{{(ak)}^2} - bkck}}{{{a^2} - bc}} = \frac{{{k^2}\left( {{a^2} - bc} \right)}}{{{a^2} - bc}} = {k^2}\)
আবার, \(\frac{{{y^2} - zx}}{{{b^2} - ca}} = \frac{{{{(bk)}^2} - ckak}}{{{b^2} - ca}} = \frac{{{k^2}\left( {{b^2} - ca} \right)}}{{{b^2} - ca}} = {k^2}\)
এবং \(\frac{{{z^2} - xy}}{{{c^2} - ab}} = \frac{{{{(ck)}^2} - ak \cdot bk}}{{{c^2} - ab}} = \frac{{{k^2}\left( {{c^2} - ab} \right)}}{{{c^2} - ab}} = {k^2}\)
সুতরাং প্রমাণিত হল যে, \(\frac{{{x^2} - yz}}{{{a^2} - bc}} = \frac{{{y^2} - zx}}{{{b^2} - ca}} = \frac{{{z^2} - xy}}{{{c^2} - ab}}\)

\(\frac{{3x + 4y}}{{3u + 4v}} = \frac{{3x - 4y}}{{3u - 4v}}\)
বা, \(\frac{{3x + 4y}}{{3x - 4y}} = \frac{{3u + 4v}}{{3u - 4v}}\)
বা, \(\frac{{3x + 4y + 3x - 4y}}{{3x + 4y - 3x + 4y}} = \frac{{3u + 4v + 3u - 4v}}{{3u + 4v - 3u + 4v}}\)(যোগ-ভাগ প্রক্রিয়া)
বা, \(\frac{{6x}}{{8y}} = \frac{{6u}}{{8v}}\)
বা, \(\frac{x}{y} = \frac{u}{v}\) [প্রমাণিত]
\(\therefore \frac{x}{y}=\frac{u}{v}\)

\({\rm{( a + b + c + d ) : ( a + b - c - d ) = ( a - b + c - d ) : ( a - b - c + d )}}\)
বা, \(\frac{{a + b + c + d}}{{a + b - c - d}} = \frac{{a - b + c - d}}{{a - b - c + d}}\)
বা, \(\frac{{a + b + c + d + a + b - c - d}}{{a + b + c + d - a - b + c + d}} = \frac{{a - b + c - d + a - b - c + d}}{{a - b + c - d - a + b + c - d}}\) (যোগ-ভাগ প্রক্রিয়া)
বা, \(\frac{{2a + 2b}}{{2c + 2d}} = \frac{{2a - 2b}}{{2c - 2d}}\)
বা, \(\frac{2(a+b)}{2(c+d)}=\frac{2(a-b)}{2(c-d)}\)
বা, \(\frac{{a + b}}{{c + d}} = \frac{{a - b}}{{c - d}}\)
বা, \((a+b)(c-d)=(a-b)(c+d)\)
বা, \({\rm{a c - a d + b c - b d = a c + a d - b c - b d}}\)
\(a c+b c-b d-a c+b c+d b=a d+a d\)
বা, \({\rm{2 b c = 2 a d}}\)
বা, \({\rm{a d = b c}}\)
বা, \(\frac{a}{b} = \frac{c}{d}\)
\(\therefore a:b = c:d\)(প্রমাণিত)

\(\frac{a^2}{b+c}=\frac{b^2}{a+a}=\frac{c^2}{a+b}=1\)
\(\therefore\) \(a^{2}=b+c, b^{2}=c+a, c^{2}=a+b\)
এখন, \(a^{2}=b+c\)
\(\therefore\) \(a+a^{2}=a+b+c\) [উভয় পক্ষের সঙ্গে \(a\) যােগ করে]
বা, \(a(1+a)=a+b+c\)
\(\therefore \frac{1}{1+a}=\frac{a}{a+b+c}\) ............ (i)
আবার, \(b^{2}=c+a\), ইহার উভয় পক্ষের সঙ্গে \(b\) যােগ করে পাই,
\(b+b^{2}=b+c+a\)
বা, \(b(1+b)=a+b+c\)
\(\therefore \frac{1}{1+b}=\frac{b}{a+b+c}\) ........... (ii)
এবং \(c^{2}=a+b\), ইহার উভয় পক্ষের সঙ্গে \(c\) যােগ করে পাই,
\(c+c^{2}=a+b+c\)
বা, \(c(1+c)=a+b+c\)
\(\frac{1}{1+c}=\frac{c}{a+b+c}\) ........... (iii)
(i), (ii), (iii) যােগ করে পাই,
\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\)
\(=\frac{a}{a+b+c}+\frac{b}{a+b+c}+\frac{c}{a+b+c}\)
\(=\frac{a+b+c}{a+b+c}=1\)
\(\therefore \frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}=1\) (প্রমাণিত)

যেহেতু, \({x^2}:(by + cz) = {y^2}:(cz + ax) = {z^2}:(ax + by) = 1\)
\(\frac{{{x^2}}}{{by + cz}} = \frac{{{y^2}}}{{cz + ax}} = \frac{{{z^2}}}{{ax + by}} = 1\)
\(\therefore {x^2} = by + cz,{y^2} = cz + ax,{z^2} = ax + by\)
এখন, \(\frac{a}{{a + x}} + \frac{b}{{b + y}} + \frac{c}{{c + z}}\)
\(= \frac{{ax}}{{(a + x)x}} + \frac{{by}}{{(b + y)y}} + \frac{{cz}}{{(c + z)z}}\)
\(=\frac{a x}{a x+x^2}+\frac{b y}{b y+y^2}+\frac{c z}{c z+z^2}\)
\( = \frac{{ax}}{{ax + by + cz}} + \frac{{by}}{{by + cz + ax}} + \frac{{cz}}{{cz + ax + by}}\) \(\left[x^2=b y+c z, y^2=c z+a x\right.\)\(\left.z^2=a x+b y\right]\)
\(= \frac{{ax + by + cz}}{{ax + by + cz}}\)
\(= 1\) (প্রমাণিত)

\(\frac{x}{x a+y b+z c}=\frac{y}{y a+zb+x c}=\frac{z}{za+x b+y c}\)
\(\because\) প্রতিটি অনুপাত \(=\frac{\text {লম্বগুলির সমষ্টি }} {\text { হরগুলির সমষ্টি }}\)
\(\therefore \quad \frac{x}{x a+y b+zc}=\frac{y}{y a+z b+x c}=\frac{z}{z a+x b+y c}\)
\(=\frac{x+y+z}{x a+y b+zc+y a+z b+x+z a+x b+y c}\) (সংযোজন প্রক্রিয়া)
\(=\frac{x+y+z}{x(a+b+c)+y(a+b+c)+z(a+b+c)}\)
\(=\frac{x+y+z}{(a+b+c)(x+y+z)}\)
\(=\frac{1}{a+b+c} \cdot[\because x+y+z \neq 0]\)
অতএব, প্রতিটি অনুপাতের মান \(\frac{1}{a+b+c}\) (প্রমাণিত)

মনে করি, \(\frac{x^{2}-y z}{a}=\frac{y^{2}-z x}{b}=\frac{z^{2}-x y}{c}=\frac{1}{k}(k \neq 0)\) ।[যেখানে k একটি সমানুপাতিক ধ্রুবক]
\(\therefore \quad a=k\left(x^{2}-y z\right), \quad b=k\left(y^{2}-z x\right), \quad c=k\left(z^{2}-x y\right)\)
এখন বামপক্ষ \(=(a+b+c)(x+y+z)\)
\(=\left\{k\left(x^{2}-y z\right)+k\left(y^{2}-z x\right)+k\left(z^{2}-x y\right)\right\}(x+y+z)\)
\(=k\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)(x+y+z)\)
\(=k\left(x^{3}+y^{3}+z^{3}-3 x y z\right)\)
এবং ডানপক্ষ \(=a x+b y+c z\)
\(=k\left(x^{2}-y z\right) x+k\left(y^{2}-z x\right) y+k\left(z^{2}-x y\right) z\)
\(=k\left(x^{3}-x y z+y^{3}-x y z+z^{3}-x y z\right)\)
\(=k\left(x^{3}+y^{3}+z^{3}-3 x y z\right)=\) বামপক্ষ।
\(\therefore\) বামপক্ষ = ডানপক্ষ (প্রমাণিত)

মনে করা যাক, \(\frac{a}{{y + z}} = \frac{b}{{z + x}} = \frac{c}{{x + y}} = k( \ne 0)\) [যেখানে k একটি সমানুপাতিক ধ্রুবক]
\(\therefore a = k(y + z),b = k(z + x),c = k(x + y)\)
এখন, \(\frac{{a(b - c)}}{{{y^2} - {z^2}}} = \frac{{k(y +z)k\{ (z + x) - (x + y)\} }}{{{y^2} - {z^2}}}\)
\(=\frac{k(y+z) k\{z+x-x-y\}}{y^2-z^2}\)
\(=\frac{-k^2-\left(y^2-z^2\right)}{y^2-z^2}=-k^2\)
আবার, \( = \frac{{{k^2}(z + y)(z - y)}}{{{y^2} - {z^2}}} = \frac{{{k^2}\left( {{z^2} - {y^2}} \right)}}{{{y^2} - {z^2}}} = - {k^2}\)
\(\frac{{b(c - a)}}{{{z^2} - {x^2}}} = \frac{{k(z + x)k\{ (x + y) - (y + z)}\}}{{{z^2} - {x^2}}}\)
\(=\frac{k(z+x) k\{x+y-y-z\}}{z^2-x^2}\)
\( = \frac{{{k^2}(x + z)(x - z)}}{{{z^2} - {x^2}}} = \frac{{{k^2}\left( {{x^2} - {z^2}} \right)}}{{{z^2} - {x^2}}} = - {k^2}\)
\(=\frac{-k^2\left(z^2-x^2\right)}{\left(z^2-x^2\right)}=-k^2\)
আবার, \(\frac{{c(a - b)}}{{{x^2} - {y^2}}} = \frac{{k(x + y)k\{ (y + z) - (z + x)\} }}{{{x^2} - {y^2}}} = \frac{{{k^2}(y + x)(y - x)}}{{{x^2} - {y^2}}} = \frac{{{k^2}\left( {{y^2} - {x^2}} \right)}}{{{x^2} - {y^2}}} = - {k^2}\)
\(=\frac{-k^2\left(x^2-y^2\right)}{\left(x^2-y^2\right)}=-k^2\)
সুতরাং প্রমাণিত হল যে, \(\frac{{a(b - c)}}{{{y^2} - {z^2}}} = \frac{{b(c - a)}}{{{z^2} - {x^2}}} = \frac{{c(a - b)}}{{{x^2} - {y^2}}}\)

চতুর্থ সমানুপাতী \( =\frac{ \text {দ্বিতীয় পদ} \times \text {তৃতীয় পদ}}{ \text {প্রথম পদ}}\)
\(\frac{{4 \times 6}}{3} = 8\)
\(\therefore\) (a) উত্তরটি ঠিক

তৃতীয় সমানুপাতী = \(\frac{{12 \times 12}}{8}\) = 18
(c) উত্তরটি ঠিক।

মধ্য সমানুপাতী = \(\sqrt{16 \times 25} = 4 \times 5=20\)

যেহেতু,\(a: \frac{27}{64}=\frac{3}{4}: a\),
সুতরাং, \(\frac{a}{\frac{27}{64}}=\frac{\frac{3}{4}}{a}\)
বা, \(a^{2}=\frac{3}{4} \times \frac{27}{64}=\frac{81}{256}\)
বা, \(a^{2}=\left(\pm \frac{9}{16}\right)^{2}\)
\(\therefore a=\pm \frac{9}{16}\)
যেহেতু, \(a\) একটি ধনাত্মক সংখ্যা, সুতরাং \(a\)-এর ধনাত্মক মানটি গ্রহণযােগ্য এবং ঋণাত্মক মানটি বর্জনীয়।
\(a\)-এর নির্ণেয় মান \(=\frac{+9}{16}\) অর্থাৎ, \(a\)-এর নির্ণেয় মান \(=\frac{9}{16}\)
উত্তর : \(a\)-এর নির্ণেয় মান \(=\frac{9}{16}\) (c)

মনেকরি, \(2a = 3b = 4c = k\) \((k \neq 0)\)[যেখানে k একটি সমানুপাতিক ধ্রুবক]
\(\therefore a = \frac{k}{2},b = \frac{k}{3},c = \frac{k}{4}\)
\(\therefore\) a : b : c
\(=\frac{k}{2}: \frac{k}{3}: \frac{k}{4}\)
\( = \frac{{k \times 6}}{{2 \times 6}}:\frac{{k \times 4}}{{3 \times 4}}:\frac{{k \times 3}}{{4 \times 3}}\)
\(=\frac{6 k}{12}: \frac{4 k}{12}: \frac{3 k}{12}\)
\(= 6k : 4k : 3k\)
\(= 6 : 4 : 3 \)
\(\therefore\) (d) উত্তরটি ঠিক।

সত্য।
কারণ, \(a b \times b c \times c a: c^2 \times a^2 \times b^2\)
\(\Rightarrow a^2 \times b^2 \times c^2: a^2 \times b^2 \times c^2\)
\(=1: 1\)

সত্য।
কারণ, \(\frac{x^3 y^2}{x^2 y^2}=\frac{x^2 y^2}{x y^3} \Rightarrow \frac{x}{y}=\frac{x}{y}\)

মনে করি, তিনটি ক্রমিক সমানুপাতী সংখ্যা যথাক্রমে \(x, y, z\)
\(\therefore \frac{x}{y}=\frac{y}{z}\)
\(\therefore y^{2}=x z \ldots\) (i)
শর্তানুসারে, \(x y z=64\)
বা, \((x z) \cdot y=64\)
বা, \(y^{2} \cdot y=64\) [(i) থেকে \(x z=y^{2}\) বসিয়ে]
বা, \(y^{3}=64=4^{3}\)
\(\therefore y=4\)
\(\therefore\) মধ্যসমানুপাতী \(= 4\).
\(\therefore\) মধ্যসমানুপাতী সংখ্যার মান \(= 4\).

12.5% ।
\(\frac{a}{2}=\frac{b}{5}=\frac{c}{8}\)
\(\Rightarrow \frac{a}{2} \times 100 \%=\frac{b}{5} \times 100 \%=\frac{c}{8} \times 100 \%\)
\(\Rightarrow a \times 50 \%=b \times 20 \%=c \times 12.5 \%\)

\(\Rightarrow-6\)
\(\sqrt{(x+2)(x-3)}=x\)
\(\Rightarrow x^2-3 x+2 x-6=x^2\)
\(\Rightarrow-x=6\)
\(\Rightarrow x=-6\)

\(\frac{a}{2} = \frac{b}{3} = \frac{c}{4} = \frac{{2a - 3b + 4c}}{p} = k\) (ধরি)\((k \neq 0)\)[যেখানে k একটি সমানুপাতিক ধ্রুবক]
\(\therefore a = 2k,b = 3k,c = 4k\)
\(\therefore 2.2 \mathrm{k}-3.3 \mathrm{k}+4.4 \mathrm{k}=\mathrm{pk} \Rightarrow 11 \mathrm{k}=\mathrm{pk} \Rightarrow \mathrm{p} = 11\)
\(\therefore\) p-এর মান \(= 11\)

\(\frac{{3x - 5y}}{{3x + 5y}} = \frac{1}{2}\)
বা, \(6 x-10 y=3 x+5 y\)
বা, \(6 x-3 y=5 y+10 y\)
বা, 3 \(x=15 y\)
বা, \(x=5 y\)
\(\therefore\) \(\frac{{3{x^2} - 5{y^2}}}{{3{x^2} + 5{y^2}}}\)
\(= \frac{{3{{(5y)}^2} - 5{y^2}}}{{3{{(5y)}^2} + 5{y^2}}}\)
\(= \frac{{75{y^2} - 5{y^2}}}{{75{y^2} + 5{y^2}}}\)
\(= \frac{{70}}{{80}}\)
\(= \frac{7}{8}\)
\(\therefore\) নির্ণেয় মান \(= \frac{7}{8}\)

ধরি \(a = 3k, b = 4k\) এবং x = 5p ও y = 7p
\(\therefore\) (3ax - by) : (4by - 7ax)
\(=(3\times 3k \times 5 p-4 k \times 7 p):(4 \times 4 k \times 7 p-7 \times 3k \times 5 p)\)
\(=(45kp-28kp):(112kp-105kp)\)\(=k P(45-28): k p(112-105)\)
\(=(45-28):(112-105)\)
\(=17: 7\)
\(\therefore\) নির্ণেয় মান \(=17: 7\)

\(\frac{x}{{12}} = \frac{{12}}{y} = \frac{y}{{27}}\)
\(\frac{{12}}{y} = \frac{y}{{27}}\)
বা, \({y^2} = 27 \times 12\)
বা, \(y = \sqrt {12 \times 27} = \sqrt {324} = 18\)
বা, \(\frac{x}{{12}} = \frac{{12}}{y}\)
বা, \(xy = 12 \times 12\)
বা, \(x = \frac{{144}}{{18}} = 8\) \([\because y=18]\)
\(\therefore\) \(x\)-এর মান 8 ও y-এর মান 18

\(a : b = 3 : 2 = 9 : 6\)
\(b : c = 3 : 2 = 6 : 4\)
\(\therefore a: b: c=9: 6: 4\)
ধরি \(a = 9k, b = 6k, c = 4k\)\((k \neq 0)\)[যেখানে k একটি সমানুপাতিক ধ্রুবক]
\(a + b : b + c = 9k + 6k : 6x + 4k\)
\(= 15k : 10k = 3 : 2\)