Mix Examples - Triangles Questions in Gujarati

(A) પ્રમેય (થેલ્સના પ્રમેય) મુજબ,જો ત્રિકોણની એક બાજુને સમાંતર રેખા બાકીની બે બાજુઓને ભિન્ન બિંદુઓમાં છેદે,તો તે બાજુઓ પર કપાતા રેખાખંડો સમપ્રમાણમાં હોય છે.
$\overline{MN} \parallel \overline{BC}$ હોવાથી,આપણી પાસે છે:
$\frac{AM}{MB} = \frac{AN}{NC}$
આપેલ છે કે $AM = 6$,$MB = 9$,અને $AN = 8$,તેથી કિંમતો મૂકતા:
$\frac{6}{9} = \frac{8}{NC}$
$\frac{2}{3} = \frac{8}{NC}$
$2 \times NC = 8 \times 3$
$2 \times NC = 24$
$NC = 12$
$AC = AN + NC$ હોવાથી:
$AC = 8 + 12 = 20$.

(B) આપેલ છે કે $\Delta ABC$ માં,$\overline{PQ} \parallel \overline{BC}$ છે.
પ્રમેય $6.1$ (થેલ્સનો પ્રમેય) મુજબ,જો ત્રિકોણની કોઈ એક બાજુને સમાંતર રેખા બાકીની બે બાજુઓને ભિન્ન બિંદુઓમાં છેદે,તો તે બાજુઓ પર કપાતા રેખાખંડો સમાન ગુણોત્તરમાં હોય છે.
તેથી,$\frac{AP}{PB} = \frac{AQ}{QC} = \frac{3}{4}$.
ધારો કે $AQ = 3x$ અને $QC = 4x$.
આપણે જાણીએ છીએ કે $AC = AQ + QC$.
કિંમતો મૂકતા,$17.5 = 3x + 4x$.
$17.5 = 7x$.
$x = \frac{17.5}{7} = 2.5$.
હવે,$AQ = 3x = 3 \times 2.5 = 7.5$.

(C) પ્રમેય (થેલ્સના પ્રમેય) મુજબ,જો ત્રિકોણની એક બાજુને સમાંતર રેખા બાકીની બે બાજુઓને ભિન્ન બિંદુઓમાં છેદે,તો તે બાજુઓ પર કપાતા રેખાખંડો તે બાજુઓનું સમાન ગુણોત્તરમાં વિભાજન કરે છે.
અહીં $\overline{MN} \parallel \overline{BC}$ હોવાથી,$\frac{AM}{MB} = \frac{AN}{NC}$ થાય.
આપણને $\frac{AM}{MB} = \frac{2}{3}$ આપેલ છે,તેથી $\frac{AN}{NC} = \frac{2}{3}$ થાય.
ધારો કે $AN = 2x$ અને $NC = 3x$.
$AC = AN + NC$ હોવાથી,$25 = 2x + 3x$ થાય.
$25 = 5x$,જેનો અર્થ છે કે $x = 5$.
તેથી,$NC = 3x = 3 \times 5 = 15$.

(D) પ્રમેય $6.1$ (થેલ્સના પ્રમેય) મુજબ,જો ત્રિકોણની કોઈ એક બાજુને સમાંતર રેખા બાકીની બે બાજુઓને ભિન્ન બિંદુઓમાં છેદે,તો તે બાજુઓ પર કપાતા રેખાખંડો સમપ્રમાણમાં હોય છે.
$\Delta XYZ$ માં,$\overline{ST} \parallel \overline{YZ}$ હોવાથી:
$\frac{XS}{SY} = \frac{XT}{TZ}$
આપેલ કિંમતો મૂકતા:
$\frac{4}{x-4} = \frac{8}{3x-19}$
ચોકડી ગુણાકાર કરતા:
$4(3x - 19) = 8(x - 4)$
$12x - 76 = 8x - 32$
$12x - 8x = 76 - 32$
$4x = 44$
$x = 11$

(A) પ્રમેય (થેલ્સના પ્રમેય) મુજબ,જો ત્રિકોણની એક બાજુને સમાંતર રેખા દોરવામાં આવે જે અન્ય બે બાજુઓને ભિન્ન બિંદુઓમાં છેદે,તો તે બાજુઓ સમાન ગુણોત્તરમાં વિભાજિત થાય છે.
આપેલ છે કે $\overline{DE} \parallel \overline{BC}$,તેથી:
$\frac{AD}{DB} = \frac{AE}{EC}$
અહીં $AD = 8$ અને $AB = 12$ આપેલ છે. $A-D-B$ હોવાથી,$DB = AB - AD = 12 - 8 = 4$ થાય.
કિંમતો મૂકતા:
$\frac{8}{4} = \frac{12}{EC}$
$2 = \frac{12}{EC}$
$EC = \frac{12}{2} = 6$.
આમ,$EC$ ની લંબાઈ $6$ છે.

(B) આપેલ છે કે $\Delta ABC$ માં,$\overline{PQ} \parallel \overline{BC}$ છે.
પ્રમેય $6.1$ (થેલ્સનો પ્રમેય અથવા પાયાનું સપ્રમાણતાનું પ્રમેય) મુજબ,જો ત્રિકોણની કોઈ એક બાજુને સમાંતર રેખા બાકીની બે બાજુઓને ભિન્ન બિંદુઓમાં છેદે,તો તે બાજુઓ પર કપાતા રેખાખંડો તે બાજુઓનું સમાન ગુણોત્તરમાં વિભાજન કરે છે.
તેથી,$\frac{AP}{AB} = \frac{AQ}{AC}$.
આપેલ કિંમતો $AP = 2$,$AB = 6$ અને $AC = 9$ છે.
આ કિંમતોને સમીકરણમાં મૂકતા:
$\frac{2}{6} = \frac{AQ}{9}$
$\frac{1}{3} = \frac{AQ}{9}$
$AQ = \frac{9}{3}$
$AQ = 3$.
આમ,$AQ$ ની કિંમત $3$ છે.

(C) પ્રમેય (થેલ્સના પ્રમેય) મુજબ,જો ત્રિકોણની એક બાજુને સમાંતર રેખા બાકીની બે બાજુઓને ભિન્ન બિંદુઓમાં છેદે,તો તે બાજુઓ પર કપાતા રેખાખંડો સમપ્રમાણમાં હોય છે.
અહીં $\overline{MN} \parallel \overline{BC}$ હોવાથી,$\frac{AM}{MB} = \frac{AN}{NC}$ થાય.
આપણને $\frac{AM}{MB} = \frac{4}{5}$ અને $NC = 2.5$ આપેલ છે.
આ કિંમતો સમીકરણમાં મૂકતા: $\frac{4}{5} = \frac{AN}{2.5}$.
બંને બાજુ $2.5$ વડે ગુણતા: $AN = \frac{4}{5} \times 2.5$.
$AN = 4 \times 0.5 = 2$.
આમ,$AN$ ની કિંમત $2$ છે.

(D) પ્રમેય $6.1$ (થેલ્સનું પ્રમેય અથવા પાયાનું સપ્રમાણતાનું પ્રમેય) મુજબ,જો ત્રિકોણની કોઈ એક બાજુને સમાંતર દોરેલી રેખા બાકીની બે બાજુઓને ભિન્ન બિંદુઓમાં છેદે,તો તે બાજુઓ પર કપાતા રેખાખંડો સપ્રમાણમાં હોય છે.
$\Delta XYZ$ માં,$\overline{ST} \parallel \overline{YZ}$ હોવાથી:
$\frac{XS}{SY} = \frac{XT}{TZ}$
અહીં $XS = 4$,$SY = 4.5$ અને $XT = 8$ આપેલ છે,તેથી:
$\frac{4}{4.5} = \frac{8}{TZ}$
$4 \times TZ = 8 \times 4.5$
$4 \times TZ = 36$
$TZ = \frac{36}{4} = 9$
$X-T-Z$ હોવાથી,$XZ = XT + TZ$ થાય.
$XZ = 8 + 9 = 17$.

(A) પ્રમેય (થેલ્સના પ્રમેય) મુજબ,જો ત્રિકોણની એક બાજુને સમાંતર રેખા બાકીની બે બાજુઓને છેદતી હોય,તો તે બાજુઓનું સમાન ગુણોત્તરમાં વિભાજન કરે છે.
આપેલ છે કે $\overline{MN} \parallel \overline{BC}$,તેથી:
$\frac{AM}{MB} = \frac{AN}{NC}$
આપેલ કિંમતો મૂકતા:
$\frac{x}{x-2} = \frac{x+2}{x-1}$
ચોકડી ગુણાકાર કરતા:
$x(x-1) = (x-2)(x+2)$
$x^2 - x = x^2 - 4$
બંને બાજુથી $x^2$ બાદ કરતા:
$-x = -4$
$x = 4$
તેથી,$x$ ની કિંમત $4$ છે.

(B) પ્રમેય $6.1$ (થેલ્સનું પ્રમેય) મુજબ,જો ત્રિકોણની એક બાજુને સમાંતર રેખા બાકીની બે બાજુઓને ભિન્ન બિંદુઓમાં છેદે,તો તે બાજુઓ પર કપાતા રેખાખંડો સમપ્રમાણમાં હોય છે.
$\overline{PQ} \parallel \overline{BC}$ હોવાથી:
$\frac{AP}{PB} = \frac{AQ}{QC}$
આપેલ કિંમતો મૂકતા:
$\frac{8x-7}{5x-3} = \frac{4x-3}{3x-1}$
ચોકડી ગુણાકાર કરતા:
$(8x-7)(3x-1) = (4x-3)(5x-3)$
$24x^2 - 8x - 21x + 7 = 20x^2 - 12x - 15x + 9$
$24x^2 - 29x + 7 = 20x^2 - 27x + 9$
પદોને એક બાજુ લાવતા:
$4x^2 - 2x - 2 = 0$
$2$ વડે ભાગતા:
$2x^2 - x - 1 = 0$
અવયવ પાડતા:
$(2x+1)(x-1) = 0$
તેથી $x = 1$ અથવા $x = -0.5$.
લંબાઈ હંમેશા ધન હોય છે,તેથી $x = 1$ મળે.

(C) પ્રમેય $6.1$ (થેલ્સના પ્રમેય) મુજબ,જો ત્રિકોણની એક બાજુને સમાંતર રેખા બાકીની બે બાજુઓને ભિન્ન બિંદુઓમાં છેદે,તો તે બાજુઓ પર કપાતા રેખાખંડો સમપ્રમાણમાં હોય છે.
$\overline{MN} \parallel \overline{BC}$ હોવાથી:
$\frac{AM}{MB} = \frac{AN}{NC}$
આપેલ કિંમતો મૂકતા:
$\frac{x-3}{2x-7} = \frac{x+3}{2x+3}$
ચોકડી ગુણાકાર કરતા:
$(x-3)(2x+3) = (x+3)(2x-7)$
$2x^2 + 3x - 6x - 9 = 2x^2 - 7x + 6x - 21$
$2x^2 - 3x - 9 = 2x^2 - x - 21$
બંને બાજુથી $2x^2$ બાદ કરતા:
$-3x - 9 = -x - 21$
પદોને ગોઠવતા:
$-3x + x = -21 + 9$
$-2x = -12$
$x = 6$
આમ,$x$ ની કિંમત $6$ છે.

(D) પ્રમેય (થેલ્સના પ્રમેય) મુજબ,જો ત્રિકોણની એક બાજુને સમાંતર રેખા બાકીની બે બાજુઓને ભિન્ન બિંદુઓમાં છેદે,તો તે બાજુઓ પર કપાતા રેખાખંડો સમપ્રમાણમાં હોય છે.
આપેલ છે કે $\overline{MN} \parallel \overline{BC}$,તેથી:
$\frac{AM}{MB} = \frac{AN}{NC}$
આપેલ કિંમતો મૂકતા:
$\frac{2.5}{3} = \frac{3.75}{NC}$
$NC = \frac{3.75 \times 3}{2.5}$
$NC = \frac{11.25}{2.5} = 4.5$
$A-N-C$ હોવાથી,$AC = AN + NC$ થાય.
$AC = 3.75 + 4.5 = 8.25$.

(A) ત્રિકોણના ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,જો કોઈ કિરણ ત્રિકોણના ખૂણાને દુભાગે,તો તે સામેની બાજુને એવા ભાગોમાં વિભાજિત કરે છે જે ત્રિકોણની અન્ય બે બાજુઓના પ્રમાણમાં હોય છે.
$\Delta ABC$ માં,$AD$ એ $\angle A$ નો દ્વિભાજક હોવાથી,આપણી પાસે ગુણોત્તર છે:
$\frac{AB}{AC} = \frac{BD}{DC}$
આપેલ કિંમતો $AB = 6,$ $BD = 4,$ અને $DC = 3$ છે.
આ કિંમતોને સૂત્રમાં મૂકતા:
$\frac{6}{AC} = \frac{4}{3}$
$AC$ શોધવા માટે ચોકડી ગુણાકાર કરતા:
$4 \times AC = 6 \times 3$
$4 \times AC = 18$
$AC = \frac{18}{4} = 4.5$
તેથી,$AC$ ની લંબાઈ $4.5$ છે.

(A) ત્રિકોણના ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,ત્રિકોણના ખૂણાનો દ્વિભાજક તેની સામેની બાજુનું એવા ભાગોમાં વિભાજન કરે છે જે ત્રિકોણની અન્ય બે બાજુઓના પ્રમાણમાં હોય છે.
તેથી,$\frac{BD}{DC} = \frac{AB}{AC}$.
આપેલ છે કે $AB = 10$ અને $AC = 14$,તેથી $\frac{BD}{DC} = \frac{10}{14} = \frac{5}{7}$.
ધારો કે $BD = 5x$ અને $DC = 7x$.
કારણ કે $BC = BD + DC = 6$,તેથી $5x + 7x = 6$,જેનો અર્થ છે કે $12x = 6$,તેથી $x = 0.5$.
આમ,$BD = 5(0.5) = 2.5$ અને $DC = 7(0.5) = 3.5$.

(C) ત્રિકોણના ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,જો કોઈ કિરણ ત્રિકોણના ખૂણાને દુભાગે છે,તો તે સામેની બાજુને એવા રેખાખંડોમાં વિભાજિત કરે છે જે ત્રિકોણની અન્ય બે બાજુઓના પ્રમાણમાં હોય છે.
$\Delta ABC$ માં,$AD$ એ $\angle A$ નો દ્વિભાજક હોવાથી,આપણી પાસે ગુણોત્તર છે:
$\frac{AB}{AC} = \frac{BD}{DC}$
આપેલ કિંમતો $AB = 5$,$AC = 4.2$ અને $BD = 2.5$ છે.
આ કિંમતોને સમીકરણમાં મૂકતા:
$\frac{5}{4.2} = \frac{2.5}{DC}$
$5 \times DC = 4.2 \times 2.5$
$5 \times DC = 10.5$
$DC = \frac{10.5}{5}$
$DC = 2.1$
આમ,$DC$ ની લંબાઈ $2.1$ છે.

(D) ત્રિકોણના ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,જો કોઈ કિરણ ત્રિકોણના ખૂણાને દુભાગે છે,તો તે સામેની બાજુને એવા ભાગોમાં વિભાજિત કરે છે જે ત્રિકોણની અન્ય બે બાજુઓના પ્રમાણમાં હોય છે.
$\Delta PQR$ માં,$PS$ એ $\angle P$ નો દ્વિભાજક હોવાથી,આપણી પાસે ગુણોત્તર છે:
$\frac{PQ}{PR} = \frac{QS}{SR}$
આપેલ કિંમતો $PQ = 5, QS = 2$ અને $SR = 3$ છે.
આ કિંમતોને સૂત્રમાં મૂકતા:
$\frac{5}{PR} = \frac{2}{3}$
ચોકડી ગુણાકાર કરતા,આપણને મળે છે:
$2 \times PR = 5 \times 3$
$2 \times PR = 15$
$PR = \frac{15}{2} = 7.5$
તેથી,$PR$ ની લંબાઈ $7.5$ છે.

(A) ત્રિકોણના ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,જો કોઈ કિરણ ત્રિકોણના ખૂણાને દુભાગે છે,તો તે સામેની બાજુને અન્ય બે બાજુઓના પ્રમાણમાં વિભાજિત કરે છે.
$\Delta XYZ$ માં,$XM$ એ $\angle X$ નો દ્વિભાજક છે,તેથી:
$\frac{XY}{XZ} = \frac{YM}{MZ}$
આપેલ છે કે $XY = 3.6$,$XZ = 4.2$ અને $MZ = 2.8$.
કિંમતો મૂકતા:
$\frac{3.6}{4.2} = \frac{YM}{2.8}$
$YM = \frac{3.6 \times 2.8}{4.2}$
$YM = \frac{3.6 \times 2}{3} = 1.2 \times 2 = 2.4$
તેથી,$YM = 2.4$.

(A) ત્રિકોણના ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,ત્રિકોણના ખૂણાનો દ્વિભાજક સામેની બાજુનું એવા ભાગોમાં વિભાજન કરે છે જે ત્રિકોણની બાકીની બે બાજુઓના પ્રમાણમાં હોય છે.
તેથી,$\frac{BD}{DC} = \frac{AB}{AC}$.
આપેલ છે કે $AB = 8$ અને $AC = 10$,તેથી $\frac{BD}{DC} = \frac{8}{10} = \frac{4}{5}$.
ધારો કે $BD = 4x$ અને $DC = 5x$.
કારણ કે $BC = BD + DC = 9$,તેથી $4x + 5x = 9$,જેનો અર્થ છે કે $9x = 9$,તેથી $x = 1$.
આમ,$BD = 4(1) = 4$ અને $DC = 5(1) = 5$.

(C) ત્રિકોણના ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,ત્રિકોણના ખૂણાનો દ્વિભાજક તેની સામેની બાજુનું એવા ભાગોમાં વિભાજન કરે છે જે ત્રિકોણની અન્ય બે બાજુઓના પ્રમાણમાં હોય છે.
$\Delta ABC$ માં,$AD$ એ $\angle A$ નો દ્વિભાજક હોવાથી,આપણી પાસે ગુણોત્તર છે: $\frac{AB}{AC} = \frac{BD}{DC}.$
આપેલ છે કે $BC = 5$ અને $DC = 3,$ તેથી $BD = BC - DC = 5 - 3 = 2.$
હવે,જાણીતી કિંમતોને ગુણોત્તરમાં મૂકતા: $\frac{AB}{4.2} = \frac{2}{3}.$
$AB$ માટે ઉકેલતા: $AB = \frac{2 \times 4.2}{3} = \frac{8.4}{3} = 2.8.$
આમ,$AB$ ની લંબાઈ $2.8$ છે.

(D) ત્રિકોણના ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,ત્રિકોણના ખૂણાનો દ્વિભાજક સામેની બાજુનું એવા ભાગમાં વિભાજન કરે છે જે ત્રિકોણની બાકીની બે બાજુઓના પ્રમાણમાં હોય છે.
$\Delta XYZ$ માં,$XM$ એ $\angle X$ નો દ્વિભાજક છે,તેથી $\frac{XY}{XZ} = \frac{YM}{MZ}$.
આપેલ છે કે $XY = 8$,$XZ = 6$ અને $MZ = 4.8$.
કિંમતો મૂકતા: $\frac{8}{6} = \frac{YM}{4.8}$.
અપૂર્ણાંકનું સાદું રૂપ આપતા: $\frac{4}{3} = \frac{YM}{4.8}$.
$YM$ માટે ઉકેલતા: $YM = \frac{4 \times 4.8}{3} = 4 \times 1.6 = 6.4$.
$YZ$ ની લંબાઈ $YM + MZ$ થાય.
$YZ = 6.4 + 4.8 = 11.2$.

(A) ત્રિકોણના ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,ત્રિકોણના ખૂણાનો દ્વિભાજક તેની સામેની બાજુનું એવા ભાગોમાં વિભાજન કરે છે જે ત્રિકોણની બાકીની બે બાજુઓના ગુણોત્તરમાં હોય છે.
$\Delta ABC$ માં,$AD$ એ $\angle A$ નો દ્વિભાજક હોવાથી,આપણને ગુણોત્તર મળે છે: $\frac{BD}{DC} = \frac{AB}{AC}$.
આપેલ છે કે $AB : AC = 3 : 4$,તેથી $\frac{BD}{DC} = \frac{3}{4}$ થાય.
ધારો કે $BD = 3x$ અને $DC = 4x$.
આપણને આપેલ છે કે $BC = 14$,તેથી $BD + DC = 14$.
કિંમતો મૂકતા,$3x + 4x = 14$,જેનું સાદું રૂપ $7x = 14$ થાય છે.
$x$ માટે ઉકેલતા,આપણને $x = 2$ મળે છે.
તેથી,$BD = 3x = 3(2) = 6$.

(B) ત્રિકોણના ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,ત્રિકોણના ખૂણાનો દ્વિભાજક સામેની બાજુનું એવા ભાગોમાં વિભાજન કરે છે જે ત્રિકોણની અન્ય બે બાજુઓના પ્રમાણમાં હોય છે.
$\Delta PQR$ માં,$PS$ એ $\angle P$ નો દ્વિભાજક હોવાથી,આપણને મળે:
$\frac{PQ}{PR} = \frac{QS}{SR}$
આપેલ છે કે $QR = 12.6$ અને $QS = 5.6$,તેથી $SR$ શોધી શકાય:
$SR = QR - QS = 12.6 - 5.6 = 7.0$
હવે,જાણીતી કિંમતોને પ્રમેયના સમીકરણમાં મૂકતા:
$\frac{8}{PR} = \frac{5.6}{7}$
$PR = \frac{8 \times 7}{5.6}$
$PR = \frac{56}{5.6} = 10$
આમ,$PR$ ની લંબાઈ $10$ છે.

(C) આપેલ છે કે $\Delta ABC$ માં $\overline{MN} \parallel \overline{BC}$ છે.
પ્રમેય $6.1$ (થેલ્સના પ્રમેય) મુજબ,જો ત્રિકોણની એક બાજુને સમાંતર રેખા બાકીની બે બાજુઓને છેદતી હોય,તો તે બાજુઓનું સમાન ગુણોત્તરમાં વિભાજન કરે છે.
તેથી,$\frac{AM}{AB} = \frac{AN}{AC}$.
આપણને $\frac{AM}{AB} = \frac{5}{7}$ અને $AN = 4$ આપેલ છે.
આ કિંમતો સમીકરણમાં મૂકતા: $\frac{5}{7} = \frac{4}{AC}$.
ચોકડી ગુણાકાર કરતા $5 \times AC = 7 \times 4$,તેથી $5 \times AC = 28$.
$AC = \frac{28}{5} = 5.6$.
$A-N-C$ હોવાથી,$AC = AN + NC$ થાય.
$5.6 = 4 + NC$.
$NC = 5.6 - 4 = 1.6$.

(D) અંતઃખંડ પ્રમેય (Intercept Theorem) મુજબ,જો ત્રણ કે તેથી વધુ સમાંતર રેખાઓને બે છેદિકાઓ દ્વારા છેદવામાં આવે,તો છેદિકાઓ પર બનતા અંતઃખંડો પ્રમાણમાં હોય છે.
તેથી,$\frac{AB}{BC} = \frac{PQ}{QR}$.
આપેલ કિંમતો $AB = 5$,$BC = 8$ અને $PQ = 6.5$ છે.
આ કિંમતોને પ્રમાણમાં મૂકતા: $\frac{5}{8} = \frac{6.5}{QR}$.
ચોકડી ગુણાકાર કરતા: $5 \times QR = 8 \times 6.5$.
$5 \times QR = 52$.
$QR = \frac{52}{5} = 10.4$.
આમ,$QR$ ની લંબાઈ $10.4$ છે.

(A) પ્રમેય મુજબ,જો ત્રણ સમાંતર રેખાઓને બે છેદિકાઓ છેદતી હોય,તો તે છેદિકાઓ પર સમાન ગુણોત્તરમાં અંતઃખંડ કાપે છે.
તેથી,પ્રથમ છેદિકા પરના અંતઃખંડોનો ગુણોત્તર એ બીજી છેદિકા પરના અનુરૂપ અંતઃખંડોના ગુણોત્તર જેટલો હોય છે:
$\frac{AB}{BC} = \frac{PQ}{QR}$
અહીં $AB = 3, BC = 4$ અને $QR = 3.6$ આપેલ છે,તેથી:
$\frac{3}{4} = \frac{PQ}{3.6}$
$PQ = \frac{3 \times 3.6}{4}$
$PQ = 3 \times 0.9 = 2.7$
આપણે $PR$ શોધવાનું છે,જ્યાં $PR = PQ + QR$ થાય.
$PR = 2.7 + 3.6 = 6.3$
આમ,$PR$ ની લંબાઈ $6.3$ છે.

(N/A) આપેલ છે: $\Delta ABC$ માં,$XM \parallel AB$ અને $XN \parallel AC$ છે. $\overline{MN}$ એ $BC$ ને $T$ માં છેદે છે.
પગલું $1$: $XM \parallel AB$ હોવાથી,$\Delta ABC$ માં પાયાના સપ્રમાણતાના પ્રમેય $(BPT)$ મુજબ,$\frac{CX}{XB} = \frac{CM}{MA}$ મળે.
પગલું $2$: $XN \parallel AC$ હોવાથી,$\Delta ABC$ માં $BPT$ મુજબ,$\frac{BX}{XC} = \frac{BN}{NA}$ મળે.
પગલું $3$: $\Delta ABC$ માં,$XM \parallel AB$ અને $XN \parallel AC$ હોવાથી,ચતુષ્કોણ $ANXM$ સમાંતરબાજુ ચતુષ્કોણ છે. તેથી,$XM = AN$ અને $XN = AM$.
પગલું $4$: $\Delta TXN$ અને $\Delta TBM$ ને ધ્યાનમાં લેતા,સમાંતર રેખાઓના ગુણધર્મનો ઉપયોગ કરતા,આપણને $\frac{TX}{TB} = \frac{TC}{TX}$ મળે છે.
પગલું $5$: ચોકડી ગુણાકાર કરતા $TX^2 = TB \times TC$ મળે છે. આમ,સાબિત થાય છે.

Given: $ABCD$ is a parallelogram. $P$ is on $BC$,and $DP$ produced meets $AB$ produced at $L$.
$(1)$ In $\triangle DCP$ and $\triangle LBP$:
$\angle DCP = \angle LBP$ (Alternate interior angles as $DC \parallel AL$)
$\angle DPC = \angle LPB$ (Vertically opposite angles)
Therefore,$\triangle DCP \sim \triangle LBP$ by $AA$ similarity.
Thus,the ratios of corresponding sides are equal: $\frac{DP}{PL} = \frac{DC}{BL} = \frac{CP}{BP}$.
Hence,$\frac{DP}{PL} = \frac{DC}{BL}$ is proved.
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(N/A) $\triangle ABC$ માં,આપણને આપેલ છે કે $\overline{PQ} \parallel \overline{BC}$ છે.
પ્રમેય $6.1$ (થેલ્સનો પ્રમેય અથવા પાયાનું સપ્રમાણતાનું પ્રમેય - $BPT$) મુજબ,જો $\overline{PQ} \parallel \overline{BC}$ હોય,તો:
$\frac{AP}{AC} = \frac{AQ}{AB}$ --- $(1)$
$\triangle ADC$ માં,આપણને આપેલ છે કે $\overline{PR} \parallel \overline{DC}$ છે.
પાયાનું સપ્રમાણતાનું પ્રમેય $(BPT)$ વાપરતા,જો $\overline{PR} \parallel \overline{DC}$ હોય,તો:
$\frac{AP}{AC} = \frac{AR}{AD}$ --- $(2)$
સમીકરણ $(1)$ અને $(2)$ પરથી,બંને પદો $\frac{AP}{AC}$ ને સમાન હોવાથી,આપણે લખી શકીએ:
$\frac{AR}{AD} = \frac{AQ}{AB}$.
આમ,સાબિત થાય છે.

(N/A) $\Delta ABC$ માં,બિંદુઓ $D$ અને $E$ એ $\overline{AB}$ પર આવેલા છે જેથી $A-D-E-B$ અને $AD = EB$ થાય.
$A-D-E-B$ હોવાથી,$AE = AD + DE$ અને $BD = DE + EB$ થાય.
આપેલ છે કે $AD = EB$,તેથી $AE = AD + DE$ અને $BD = DE + AD$,એટલે કે $AE = BD$ ... $(1)$.
$\Delta ABC$ માં,$\overline{DP} \parallel \overline{BC}$ હોવાથી,થેલ્સના પ્રમેય $(BPT)$ મુજબ:
$\frac{AD}{DB} = \frac{AP}{PC}$ ... $(2)$.
$\Delta ABC$ માં,$\overline{EQ} \parallel \overline{AC}$ હોવાથી,થેલ્સના પ્રમેય $(BPT)$ મુજબ:
$\frac{BE}{EA} = \frac{BQ}{QC}$ થાય.
$AE = BD$ અને $AD = EB$ હોવાથી,આ સમીકરણ $\frac{AD}{BD} = \frac{BQ}{QC}$ બને છે ... $(3)$.
સમીકરણ $(2)$ અને $(3)$ પરથી:
$\frac{AP}{PC} = \frac{BQ}{QC}$ મળે છે.
$\Delta ABC$ માં થેલ્સના પ્રમેયના પ્રતિપ પ્રમેય મુજબ,જો $\frac{AP}{PC} = \frac{BQ}{QC}$ હોય,તો $\overline{PQ} \parallel \overline{AB}$ સાબિત થાય છે.

(N/A) આપેલ છે: $\Delta ABC$ માં,$A-D-B$,$A-E-C$,$BD = CE$ અને $m\angle B = m\angle C$.
પગલું $1$: $m\angle B = m\angle C$ હોવાથી,સમાન ખૂણાઓની સામેની બાજુઓ સમાન હોય છે. તેથી,$AB = AC$.
પગલું $2$: આપણે બાજુઓને $AB = AD + DB$ અને $AC = AE + EC$ તરીકે દર્શાવી શકીએ છીએ (કારણ કે $A-D-B$ અને $A-E-C$ છે).
પગલું $3$: આ કિંમતોને $AB = AC$ માં મૂકતા,આપણને $AD + DB = AE + EC$ મળે છે.
પગલું $4$: $BD = CE$ હોવાથી,આપણે સમીકરણમાં $CE$ ની જગ્યાએ $BD$ મૂકી શકીએ છીએ: $AD + DB = AE + DB$.
પગલું $5$: બંને બાજુથી $DB$ બાદ કરતા,આપણને $AD = AE$ મળે છે.
પગલું $6$: હવે,ગુણોત્તર $\frac{AD}{DB}$ અને $\frac{AE}{EC}$ ને ધ્યાનમાં લો. $AD = AE$ અને $DB = EC$ હોવાથી,$\frac{AD}{DB} = \frac{AE}{EC}$ થાય છે.
પગલું $7$: પાયાના સપ્રમાણતાના પ્રમેયના પ્રતિપ (થેલ્સના પ્રમેય) મુજબ,જો કોઈ રેખા ત્રિકોણની બે બાજુઓનું સમાન ગુણોત્તરમાં વિભાજન કરે,તો તે રેખા ત્રીજી બાજુને સમાંતર હોય છે. તેથી,$\overline{DE} \parallel \overline{BC}$.

(N/A) આપેલ છે: બિંદુ $O$ એ $\Delta PQR$ ના અંદરના ભાગમાં આવેલું છે. $O-A-P$,$O-B-Q$,$O-C-R$,$\overline{AB} \parallel \overline{PQ}$ અને $\overline{BC} \parallel \overline{QR}$.
સાબિત કરવાનું છે: $\overline{AC} \parallel \overline{PR}$.
સાબિતી:
$\Delta OPQ$ માં,$\overline{AB} \parallel \overline{PQ}$ હોવાથી,થેલ્સના પ્રમેય $(BPT)$ મુજબ:
$\frac{OA}{AP} = \frac{OB}{BQ} \quad \dots (1)$
$\Delta OQR$ માં,$\overline{BC} \parallel \overline{QR}$ હોવાથી,થેલ્સના પ્રમેય $(BPT)$ મુજબ:
$\frac{OB}{BQ} = \frac{OC}{CR} \quad \dots (2)$
પરિણામ $(1)$ અને $(2)$ પરથી:
$\frac{OA}{AP} = \frac{OC}{CR}$
$\Delta OPR$ માં,$\frac{OA}{AP} = \frac{OC}{CR}$ હોવાથી,થેલ્સના પ્રમેયના પ્રતિપ્રમેય મુજબ:
$\overline{AC} \parallel \overline{PR}$.

(N/A) $\triangle ABC$ માં,$AE$ એ $\angle BAC$ નો ખૂણાનો દ્વિભાજક છે. ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,આપણી પાસે $\frac{BE}{EC} = \frac{AB}{AC}$ છે.
$\triangle ADC$ માં,$AF$ એ $\angle DAC$ નો ખૂણાનો દ્વિભાજક છે. ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,આપણી પાસે $\frac{DF}{FC} = \frac{AD}{AC}$ છે.
આપેલ છે કે $AB = AD$,તેથી બીજા સમીકરણમાં $AD$ ની જગ્યાએ $AB$ મૂકતા: $\frac{DF}{FC} = \frac{AB}{AC}$.
બંને સમીકરણોની સરખામણી કરતા,આપણને $\frac{BE}{EC} = \frac{DF}{FC}$ મળે છે.
$\triangle BCD$ માં,પાયાના સપ્રમાણતાના પ્રમેય (થેલ્સના પ્રમેય) ના પ્રતિપ મુજબ,$\frac{BE}{EC} = \frac{DF}{FC}$ હોવાથી,તે સાબિત થાય છે કે $\overline{EF} \parallel \overline{BD}$.

(A) $\Delta AOB$ માં,$OD$ એ $\angle AOB$ નો ખૂણાનો દ્વિભાજક છે. ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,ખૂણાની સામેની બાજુના રેખાખંડોનો ગુણોત્તર એ ત્રિકોણની અન્ય બે બાજુઓના ગુણોત્તર જેટલો હોય છે.
તેથી,$\frac{AD}{DB} = \frac{OA}{OB}$.
તે જ રીતે,$\Delta BOC$ માં,$OE$ એ $\angle BOC$ નો ખૂણાનો દ્વિભાજક છે. તેથી,$\frac{BE}{EC} = \frac{OB}{OC}$.
$\Delta COA$ માં,$OF$ એ $\angle COA$ નો ખૂણાનો દ્વિભાજક છે. તેથી,$\frac{CF}{FA} = \frac{OC}{OA}$.
આ ત્રણેય સમીકરણોનો ગુણાકાર કરતા:
$\frac{AD}{DB} \times \frac{BE}{EC} \times \frac{CF}{FA} = \frac{OA}{OB} \times \frac{OB}{OC} \times \frac{OC}{OA}$.
$\frac{AD \times BE \times CF}{DB \times EC \times FA} = 1$.
તેથી,$AD \times BE \times CF = DB \times EC \times FA$.

(N/A) $\Delta ABD$ માં,$DE$ એ $\angle ADB$ નો દ્વિભાજક છે. ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,$\frac{AE}{EB} = \frac{AD}{DB}$.
$\overline{AD}$ મધ્યગા હોવાથી,$D$ એ $\overline{BC}$ નું મધ્યબિંદુ છે,તેથી $DB = DC$. આમ,$\frac{AE}{EB} = \frac{AD}{DC}$.
$\Delta ADC$ માં,$DF$ એ $\angle ADC$ નો દ્વિભાજક છે. ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,$\frac{AF}{FC} = \frac{AD}{DC}$.
બંને સમીકરણોની સરખામણી કરતા,આપણને $\frac{AE}{EB} = \frac{AF}{FC}$ મળે છે.
$\Delta ABC$ માં પાયાના સપ્રમાણતાના પ્રમેય (થેલ્સનો પ્રમેય) ના પ્રતિપ મુજબ,$\frac{AE}{EB} = \frac{AF}{FC}$ હોવાથી,$\overline{EF} \parallel \overline{BC}$ સાબિત થાય છે.

(N/A) ધારો કે $\triangle ABC$ એક ત્રિકોણ છે જેમાં $AD$ એ $\angle A$ નો ખૂણાનો દ્વિભાજક છે,જેથી $D$ એ $BC$ પર આવેલ છે અને $BD = DC$ છે.
$\triangle ABD$ અને $\triangle ACD$ માં:
$1$. $\angle BAD = \angle CAD$ (કારણ કે $AD$ એ $\angle A$ નો દ્વિભાજક છે).
$2$. $AD = AD$ (સામાન્ય બાજુ).
$3$. $BD = DC$ (આપેલ છે).
બાજુ-ખૂણો-બાજુ $(SAS)$ એકરૂપતાની શરત મુજબ,$\triangle ABD \cong \triangle ACD$.
ત્રિકોણો એકરૂપ હોવાથી,તેમના અનુરૂપ ભાગો સમાન હોય છે $(CPCT)$.
તેથી,$AB = AC$.
ત્રિકોણની બે બાજુઓ સમાન હોવાથી,$\triangle ABC$ એ સમદ્વિબાજુ ત્રિકોણ છે.

(N/A) આપેલ છે: $\overline{BD}$ એ $\angle B$ નો દ્વિભાજક છે,તેથી $\angle EBD = \angle DBC = \frac{1}{2} \angle B$.
$\overline{CE}$ એ $\angle C$ નો દ્વિભાજક છે,તેથી $\angle DCE = \angle ECB = \frac{1}{2} \angle C$.
કારણ કે $\overline{DE} \parallel \overline{BC}$,યુગ્મકોણની જોડ સમાન હોવાથી,$\angle EDB = \angle DBC$.
આમ,$\angle EDB = \angle EBD$,જે દર્શાવે છે કે $\Delta EBD$ સમদ্বિબાજુ છે અને $EB = ED$.
તે જ રીતે,$\angle DEC = \angle ECB$,તેથી $\angle DEC = \angle DCE$,જે દર્શાવે છે કે $\Delta DCE$ સમদ্বિબાજુ છે અને $DC = ED$.
તેથી,$EB = DC$.
$\Delta ABC$ માં,$\overline{DE} \parallel \overline{BC}$ હોવાથી,પાયાના સપ્રમાણતાના પ્રમેય (થેલ્સનું પ્રમેય) મુજબ,$\frac{AE}{EB} = \frac{AD}{DC}$.
$EB = DC$ હોવાથી,આપણને $AE = AD$ મળે છે.
બંને બાજુ $EB$ ઉમેરતા: $AE + EB = AD + DC$,જે $AB = AC$ આપે છે.
બે બાજુઓ સમાન હોવાથી,$\Delta ABC$ એ સમদ্বિબાજુ ત્રિકોણ છે.

(N/A) $1$. $\Delta ABC$ માં,ધારો કે $BO$ એ $\angle B$ નો દ્વિભાજક છે અને $CO$ એ $\angle C$ નો દ્વિભાજક છે. $O$ એ આ દ્વિભાજકોનું છેદબિંદુ હોવાથી,ત્રિકોણના અંતઃકેન્દ્રના ગુણધર્મ મુજબ $AO$ એ $\angle A$ નો દ્વિભાજક છે.
$2$. ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,$\Delta ABC$ માં,જો $AP$ એ $\angle A$ નો દ્વિભાજક હોય જે $BC$ ને $P$ માં છેદે છે,તો ખૂણો બનાવતી બાજુઓનો ગુણોત્તર એ સામેની બાજુના રેખાખંડોના ગુણોત્તર જેટલો હોય છે.
$3$. તેથી,$\frac{AB}{AC} = \frac{BP}{PC}$.
$4$. આ $\Delta ABC$ માટે ખૂણાના દ્વિભાજકના પ્રમેયની પુષ્ટિ કરે છે જ્યાં $AP$ એ $\angle A$ નો આંતરિક ખૂણાનો દ્વિભાજક છે.

(N/A) ધારો કે $ABCD$ એક બહિર્મુખ ચતુષ્કોણ છે. ધારો કે $P, Q, R,$ અને $S$ એ અનુક્રમે બાજુઓ $AB, BC, CD,$ અને $DA$ ના મધ્યબિંદુઓ છે.
$AC$ ને જોડો. $\triangle ABC$ માં,$P$ અને $Q$ એ $AB$ અને $BC$ ના મધ્યબિંદુઓ છે. મધ્યબિંદુ પ્રમેય મુજબ,$PQ \parallel AC$ અને $PQ = \frac{1}{2} AC$ (સમીકરણ $1$).
$\triangle ADC$ માં,$S$ અને $R$ એ $AD$ અને $CD$ ના મધ્યબિંદુઓ છે. મધ્યબિંદુ પ્રમેય મુજબ,$SR \parallel AC$ અને $SR = \frac{1}{2} AC$ (સમીકરણ $2$).
સમીકરણ $1$ અને સમીકરણ $2$ પરથી,આપણને $PQ \parallel SR$ અને $PQ = SR$ મળે છે.
ચતુષ્કોણ $PQRS$ ની સામસામેની બાજુઓની એક જોડ સમાંતર અને સમાન હોવાથી,$PQRS$ એક સમાંતરબાજુ ચતુષ્કોણ છે.

(N/A) $1$. $\triangle ABD$ માં,આપણને $\frac{AP}{PB} = \frac{AS}{SD}$ આપેલ છે. પાયાના સપ્રમાણતાના પ્રમેય (થેલ્સના પ્રમેય) ના પ્રતિપ વિધાન મુજબ,$PS \parallel BD$ થાય.
$2$. $\triangle BCD$ માં,આપણને $\frac{CB}{QB} = \frac{CR}{RD}$ આપેલ છે. આને $\frac{BQ}{QC} = \frac{DR}{RC}$ તરીકે ફરીથી લખી શકાય છે. પાયાના સપ્રમાણતાના પ્રમેયના પ્રતિપ વિધાન મુજબ,$QR \parallel BD$ થાય.
$3$. કારણ કે $PS \parallel BD$ અને $QR \parallel BD$,અને એક જ રેખાને સમાંતર રેખાઓ પરસ્પર સમાંતર હોય છે,તેથી આપણે કહી શકીએ કે $PS \parallel QR$.

(N/A) $1$. $\Delta ADB$ માં,$DP$ એ $\angle ADB$ નો દ્વિભાજક છે. ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,$\frac{AP}{PB} = \frac{AD}{BD}$,જેનો અર્થ છે કે $AP = \frac{AD \times PB}{BD}$.
$2$. $\Delta ADC$ માં,$DQ$ એ $\angle ADC$ નો દ્વિભાજક છે. ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,$\frac{AQ}{QC} = \frac{AD}{DC}$,જેનો અર્થ છે કે $AQ = \frac{AD \times QC}{DC}$.
$3$. આ બંને સમીકરણોનો ગુણાકાર કરતા: $AP \times AQ = \frac{AD \times PB}{BD} \times \frac{AD \times QC}{DC} = \frac{AD^2 \times PB \times QC}{BD \times DC}$.
$4$. પુનઃગોઠવણી કરતા $AP \times AQ \times BD \times DC = AD^2 \times PB \times QC$ મળે છે.
$5$. જો $\overleftrightarrow{PQ} \parallel \overleftrightarrow{BC}$ હોય,તો $\Delta ABC$ માં પાયાના સપ્રમાણતાના પ્રમેય $(BPT)$ મુજબ,$\frac{AP}{PB} = \frac{AQ}{QC}$ થાય.
$6$. ખૂણાના દ્વિભાજકના પ્રમેય મુજબ,$\frac{AP}{PB} = \frac{AD}{BD}$ અને $\frac{AQ}{QC} = \frac{AD}{DC}$ છે.
$7$. તેથી,$\frac{AD}{BD} = \frac{AD}{DC}$,જેનો અર્થ છે કે $BD = DC$. આમ,$D$ એ $\overline{BC}$ નું મધ્યબિંદુ છે.

(A) ધારો કે ત્રિકોણના શિરોબિંદુઓ $A(x_1, y_1)$,$B(x_2, y_2)$ અને $C(x_3, y_3)$ છે.
મધ્યકેન્દ્ર $G = (\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3})$ દ્વારા આપવામાં આવે છે.
આપેલ છે કે $3 AN = 2 AC$,તેથી $\frac{AN}{AC} = \frac{2}{3}$,જેનો અર્થ છે કે $N$ એ $AC$ નું $2:1$ ગુણોત્તરમાં વિભાજન કરે છે.
વિભાજન સૂત્રનો ઉપયોગ કરીને,$N$ ના યામ $(\frac{2x_3+x_1}{3}, \frac{2y_3+y_1}{3})$ મળે છે.
રેખા $MN$ એ $G$ માંથી પસાર થતી હોવાથી,બિંદુઓ $M, G, N$ સમરેખ છે.
મધ્યકેન્દ્રના ગુણધર્મ અને રેખાખંડોના ગુણોત્તરનો ઉપયોગ કરીને,$\Delta ABC$ માં છેદિકા $MGN$ માટે મેનેલાઉસના પ્રમેય અથવા સદિશ ભૂમિતિ દ્વારા,આપણે $AM:MB$ નો ગુણોત્તર મેળવી શકીએ છીએ.
શરત $3 AN = 2 AC$ મુજબ,$N$ એવું બિંદુ છે કે જેથી $AN = \frac{2}{3} AC$ થાય.
મધ્યકેન્દ્રમાંથી પસાર થતી છેદિકાઓના ગુણધર્મનો ઉપયોગ કરતા,આપણને $\frac{AM}{MB} = 2$ મળે છે,જેનો અર્થ છે કે $AM = 2 MB$.

(A) આપેલ છે કે $\square ABCD$ એક સમલંબ ચતુષ્કોણ છે જેમાં $\overline{AB} \parallel \overline{CD}$ છે.
આપણને $AM \times NC = BN \times MD$ આપેલ છે,જેને $\frac{AM}{MD} = \frac{BN}{NC}$ તરીકે લખી શકાય.
કારણ કે $\frac{AM}{MD} = \frac{2}{3}$,તેથી $\frac{BN}{NC} = \frac{2}{3}$ થાય.
પ્રમેય (થેલ્સના પ્રમેય) મુજબ,જો $\frac{AM}{MD} = \frac{BN}{NC}$ હોય,તો રેખાખંડ $\overline{MN}$ એ પાયા $\overline{AB}$ અને $\overline{CD}$ ને સમાંતર હોય.
ધારો કે $\overline{MN} \parallel \overline{AB} \parallel \overline{CD}$.
$\triangle ADC$ માં,$\overline{MO} \parallel \overline{CD}$ હોવાથી,સમરૂપ ત્રિકોણની શરત $(\triangle AMO \sim \triangle ADC)$ મુજબ,$\frac{AO}{AC} = \frac{AM}{AD}$ મળે.
આપેલ છે કે $\frac{AM}{MD} = \frac{2}{3}$,ધારો કે $AM = 2k$ અને $MD = 3k$. તેથી $AD = AM + MD = 2k + 3k = 5k$.
આમ,$\frac{AO}{AC} = \frac{AM}{AD} = \frac{2k}{5k} = \frac{2}{5}$.

(C) આપેલ છે કે $\Delta ABC$ અને $\Delta PQR$ માં,$\angle A = \angle P$ અને $\angle B = \angle R$ છે.
$AA$ (ખૂ-ખૂ) સમરૂપતાની શરત મુજબ,$\Delta ABC \sim \Delta PRQ$ થાય.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે:
$\frac{AB}{PR} = \frac{AC}{PQ}$
આપેલ કિંમતો મૂકતા: $AB = 8$,$PQ = 7.5$,$AC = 6$.
$\frac{8}{PR} = \frac{6}{7.5}$
$PR = \frac{8 \times 7.5}{6}$
$PR = \frac{60}{6} = 10$.
તેથી,$PR = 10$.

(D) $\Delta ABC$ માં,$A-M-B$,$A-N-C$ અને $\overline{MN} \parallel \overline{BC}$ છે.
$\overline{MN} \parallel \overline{BC}$ હોવાથી,$\angle AMN \cong \angle ABC$ અને $\angle ANM \cong \angle ACB$ (અનુકોણ).
$AA$ સમરૂપતાની શરત મુજબ,$\Delta AMN \sim \Delta ABC$.
તેથી,અનુરૂપ બાજુઓનો ગુણોત્તર સમાન થાય:
$\frac{AM}{AB} = \frac{MN}{BC}$
આપણે જાણીએ છીએ કે $AB = AM + MB = 2 + 3 = 5$.
કિંમતો મૂકતા:
$\frac{2}{5} = \frac{3}{BC}$
$2 \times BC = 3 \times 5$
$2 \times BC = 15$
$BC = \frac{15}{2} = 7.5$

(A) આપેલ છે કે $\Delta ABC$ અને $\Delta XYZ$ માં,$\frac{AB}{XY} = \frac{BC}{YZ}$ અને $\angle B \cong \angle Y$ છે.
$SAS$ (બાજુ-ખૂણો-બાજુ) સમરૂપતાની શરત મુજબ,$\Delta ABC \sim \Delta XYZ$ થાય.
ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય છે:
$\frac{AB}{XY} = \frac{AC}{XZ} = \frac{BC}{YZ}$.
આપેલ કિંમતો મૂકતા:
$\frac{6}{15} = \frac{15}{XZ}$.
$XZ$ શોધવા માટે ચોકડી ગુણાકાર કરતા:
$6 \times XZ = 15 \times 15$.
$6 \times XZ = 225$.
$XZ = \frac{225}{6} = 37.5$.
તેથી,$XZ = 37.5$.

(C) સમરૂપતા નક્કી કરવા માટે,આપણે $\Delta ABC$ અને $\Delta PQR$ ની અનુરૂપ બાજુઓના ગુણોત્તરની સરખામણી કરીએ છીએ.
પ્રથમ,દરેક ત્રિકોણની બાજુઓને વધતા ક્રમમાં ગોઠવો:
$\Delta ABC$ માટે: $AB = 3$,$BC = 6$,$AC = 8$.
$\Delta PQR$ માટે: $PR = 5.1$,$QR = 10.2$,$PQ = 13.6$.
હવે,અનુરૂપ બાજુઓના ગુણોત્તરની ગણતરી કરો:
સૌથી નાની બાજુઓનો ગુણોત્તર: $\frac{AB}{PR} = \frac{3}{5.1} = \frac{30}{51} = \frac{10}{17}$.
મધ્યમ બાજુઓનો ગુણોત્તર: $\frac{BC}{QR} = \frac{6}{10.2} = \frac{60}{102} = \frac{10}{17}$.
સૌથી મોટી બાજુઓનો ગુણોત્તર: $\frac{AC}{PQ} = \frac{8}{13.6} = \frac{80}{136} = \frac{10}{17}$.
અહીં $\frac{AB}{PR} = \frac{BC}{QR} = \frac{AC}{PQ} = \frac{10}{17}$ હોવાથી,$SSS$ સમરૂપતાની શરત મુજબ,$ABC \leftrightarrow PRQ$ સંગતતા સમરૂપતા છે.

(C) આપેલ છે કે સંગતતા $PQR \leftrightarrow XZY$ માટે $\Delta PQR \sim \Delta XYZ$ છે.
સમરૂપ ત્રિકોણના ક્ષેત્રફળના પ્રમેય મુજબ,બે સમરૂપ ત્રિકોણના ક્ષેત્રફળનો ગુણોત્તર તેમની અનુરૂપ બાજુઓના વર્ગના ગુણોત્તર જેટલો હોય છે.
તેથી,$\frac{\Delta PQR \text{ નું ક્ષેત્રફળ}}{\Delta XYZ \text{ નું ક્ષેત્રફળ}} = \frac{PQ^2}{XZ^2}$.
આપેલ કિંમતો મૂકતા: $\frac{24}{\Delta XYZ \text{ નું ક્ષેત્રફળ}} = \frac{8^2}{12^2}$.
$\frac{24}{\Delta XYZ \text{ નું ક્ષેત્રફળ}} = \frac{64}{144}$.
અપૂર્ણાંકનું સાદું રૂપ આપતા: $\frac{64}{144} = \frac{4}{9}$.
તેથી,$\frac{24}{\Delta XYZ \text{ નું ક્ષેત્રફળ}} = \frac{4}{9}$.
$\Delta XYZ \text{ નું ક્ષેત્રફળ} = \frac{24 \times 9}{4} = 6 \times 9 = 54$.

(N/A) આપેલ છે: $\Delta ABC \sim \Delta PQR$ સંગતતા $ABC \leftrightarrow PQR$ માટે. $\overline{AD}$ અને $\overline{PM}$ એ અનુક્રમે $\Delta ABC$ અને $\Delta PQR$ ની મધ્યગાઓ છે.
સાબિત કરવાનું છે: $\frac{AD}{PM} = \frac{AB}{PQ}.$
સાબિતી:
$1$. $\Delta ABC \sim \Delta PQR$ હોવાથી,તેમની અનુરૂપ બાજુઓ સમપ્રમાણમાં હોય અને અનુરૂપ ખૂણાઓ એકરૂપ હોય.
$\therefore \frac{AB}{PQ} = \frac{BC}{QR} = \frac{AC}{PR}$ અને $\angle B \cong \angle Q \quad \dots (1)$
$2$. $\overline{AD}$ અને $\overline{PM}$ મધ્યગાઓ હોવાથી,$D$ એ $BC$ નું મધ્યબિંદુ છે અને $M$ એ $QR$ નું મધ્યબિંદુ છે.
તેથી,$BC = 2BD$ અને $QR = 2QM.$
$3$. આ કિંમતો સમીકરણ $(1)$ માં મૂકતા:
$\frac{AB}{PQ} = \frac{2BD}{2QM} = \frac{BD}{QM} \quad \dots (2)$
$4$. $\Delta ABD$ અને $\Delta PQM$ માં:
સમીકરણ $(1)$ પરથી,$\angle B \cong \angle Q.$
સમીકરણ $(2)$ પરથી,$\frac{AB}{PQ} = \frac{BD}{QM}.$
$5$. $SAS$ (બાખૂબા) સમરૂપતાની શરત મુજબ,$\Delta ABD \sim \Delta PQM.$
$6$. ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓ સમપ્રમાણમાં હોય:
$\therefore \frac{AD}{PM} = \frac{AB}{PQ}.$ આમ સાબિત થાય છે.

(N/A) આપેલ છે: $\Delta ABC \sim \Delta XYZ$ સંગતતા $ABC \leftrightarrow XYZ$ માટે. $\overline{AD} \perp \overline{BC}$ અને $\overline{XM} \perp \overline{YZ}$.
સાબિત કરવાનું છે: $\frac{AD}{XM} = \frac{BC}{YZ}$.
સાબિતી:
$1$. $\Delta ABC \sim \Delta XYZ$ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય અને અનુરૂપ ખૂણાઓ એકરૂપ હોય.
$\therefore \frac{AB}{XY} = \frac{BC}{YZ} = \frac{AC}{XZ}$ અને $\angle B \cong \angle Y$ ... $(1)$
$2$. $\Delta ABD$ અને $\Delta XYM$ માં:
$\angle B \cong \angle Y$ (પરિણામ $1$ પરથી)
$\angle ADB \cong \angle XMY$ (બંને $90^{\circ}$ છે કારણ કે $\overline{AD} \perp \overline{BC}$ અને $\overline{XM} \perp \overline{YZ}$)
$3$. $AA$ સમરૂપતાની શરત મુજબ,$\Delta ABD \sim \Delta XYM$.
$4$. ત્રિકોણો સમરૂપ હોવાથી,તેમની અનુરૂપ બાજુઓનો ગુણોત્તર સમાન હોય:
$\therefore \frac{AD}{XM} = \frac{AB}{XY}$ ... $(2)$
$5$. પરિણામ $(1)$ અને $(2)$ પરથી,$\frac{AB}{XY} = \frac{BC}{YZ}$ હોવાથી,આપણને મળે:
$\frac{AD}{XM} = \frac{BC}{YZ}$.
આમ,સાબિત થાય છે.

(A) $\Delta ABC$ માં,$m\angle B = 90^\circ$ અને $\overline{BM} \perp \overline{AC}$ છે.
$(i)$ સંગતતા $AMB \leftrightarrow ABC$ માટે:
$\angle AMB \cong \angle ABC$ (બંને $90^\circ$ છે)
$\angle MAB \cong \angle BAC$ (સામાન્ય ખૂણો)
તેથી,$AA$ સમરૂપતાની શરત મુજબ,$\Delta AMB \sim \Delta ABC$.
(ii) સંગતતા $BMC \leftrightarrow ABC$ માટે:
$\angle BMC \cong \angle ABC$ (બંને $90^\circ$ છે)
$\angle MCB \cong \angle BCA$ (સામાન્ય ખૂણો)
તેથી,$AA$ સમરૂપતાની શરત મુજબ,$\Delta BMC \sim \Delta ABC$.
(iii) કારણ કે $\Delta AMB \sim \Delta ABC$ અને $\Delta BMC \sim \Delta ABC$,સમરૂપતાના પરંપરિતતાના ગુણધર્મ મુજબ,$\Delta AMB \sim \Delta BMC$.