Plzzzzzzz... help......

1.	Plzzzzzzz... help......
Answer» ɢɪᴠᴇɴ :- An equilateral triangle ∆ABC . ᴛᴏ ᴘʀᴏᴠᴇ :- An equilateral triangle is equiangular . ᴄᴏɴsᴛʀᴜᴄᴛɪᴏɴ:- From vertex A drop a perpendicular on side BC . Mark it as point M . From vertex A drop a perpendicular on side BC and mark it as point N . ᴘʀᴏᴏғ :- Firstly we know that in an equilateral triangle perpendicular bisector , altitude and median all concide with each other . Also they are equal for each of the vertices . $\underline{\textsf{\textbf{\red{\purple{$\leadsto$}\:Figure:-}}}}$ $\setlength{\unitlength}{1 cm}\begin{picture}(12,8)\put(0,0){\line(1,2){2}}\put(2,4){\line(0,-1){4}}\put(4,0){\line(-1,2){2}}\put(0,0){\line(1,0){4}}\put(1.2,1.8){\line(-1,1){0.36}}\put(2.9,1.8){\line(1,1){0.36}}\put(1,0.3){\line(0,-1){0.6}}\put(1.2,0.3){\line(0,-1){0.6}}\put(3,0.3){\line(0,-1){0.6}}\put(3.2,0.3){\line(0,-1){0.6}}\put(2.2,0){\line(0,1){0.3}}\put(2.2,0.3){\line(-1,0){0.25}}\put(2,-0.4){$\bf M $}\put(0,-0.4){$ \bf A $ }\put(4,-0.4){$\bf <klux>B</klux> $ }\put(2,4.3){$ \bf C $ }\qbezier(0.5,0)(0.4,0.9)(0.3,0.65)\qbezier(1.8,3.5)(2,3)(2.3,3.5)\qbezier(3.5,0)(3.6,0.5)(3.7,0.65)\put(3,2){\line(-3,-2){3}}\put(3.2,1.7){\line(-3,-2){0.36}}\put(2.9,1.45){\line(-1,2){0.18}}\put(3.3,2.3){$ \bf N $}\end{picture}$ ____________________________________ Now we are required to PROVE that ∠ A = ∠ B = ∠ C , that is all angles are equal and each equals to 60°. So , by construction we can SAY that ∠ AMC = ∠BMC = 90° and AM = BM since CM is perpendicular bisector . $\underline{\textsf{\textbf{\green{\orange{$\mapsto$} In\:$\triangle$\:AMC \:and\:$\triangle$\:BMC}}}}$ $\tt{\blue\qquad\qquad\pink{\bullet}\:AM=BM(By\: Construction)}$ $\tt{\blue\qquad\qquad\pink{\bullet}\:CM=CM(Common)}$ $\tt{\blue\qquad\qquad\pink{\bullet}\:\angle AMC=\angle BMC (By\: Construction)}$ $\sf \therefore By \:SAS\: congruence\: condition,\\\sf \qquad\red{\triangle AMC\cong \triangle BMC}$ Therefore by corresponding parts of congruent triangles we can say that $\sf \blue{AC = BC}$ $\sf\blue {\angle ACM = \angle BCM}$ $\sf\blue {\angle CAM= \angle CBM}$ ___________________________________ $\underline{\textsf{\textbf{\green{\orange{$\mapsto$} In\:$\triangle$\:ACN \:and\:$\triangle$\:ABN}}}}$ SIMILARLY we can prove that ∆ BNA $\cong$ ∆ CNA . Therefore by corresponding parts of congruent triangles we can say that $\sf \blue{ AB = AC}$ $\sf\blue {\angle ABN = \angle ACN}$ $\sf\blue {\angle BAN= \angle CAN}$ From all these we can conclude that $\qquad\qquad\bf{\orange{\green{\bullet}\:\angle ABC=\angle ACB=\angle BAC}}$ $\boxed{\purple{\sf\orange{\leadsto}\:Hence\:an\: equilateral\: triangle\:is\: equiangular.}}$

Answer»

ɢɪᴠᴇɴ :-

An equilateral triangle ∆ABC .

ᴛᴏ ᴘʀᴏᴠᴇ :-

An equilateral triangle is equiangular .

ᴄᴏɴsᴛʀᴜᴄᴛɪᴏɴ:-

From vertex A drop a perpendicular on side BC . Mark it as point M .

From vertex A drop a perpendicular on side BC and mark it as point N .

ᴘʀᴏᴏғ :-

Firstly we know that in an equilateral triangle perpendicular bisector , altitude and median all concide with each other . Also they are equal for each of the vertices .

$\underline{\textsf{\textbf{\red{\purple{$\leadsto$}\:Figure:-}}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(12,8)\put(0,0){\line(1,2){2}}\put(2,4){\line(0,-1){4}}\put(4,0){\line(-1,2){2}}\put(0,0){\line(1,0){4}}\put(1.2,1.8){\line(-1,1){0.36}}\put(2.9,1.8){\line(1,1){0.36}}\put(1,0.3){\line(0,-1){0.6}}\put(1.2,0.3){\line(0,-1){0.6}}\put(3,0.3){\line(0,-1){0.6}}\put(3.2,0.3){\line(0,-1){0.6}}\put(2.2,0){\line(0,1){0.3}}\put(2.2,0.3){\line(-1,0){0.25}}\put(2,-0.4){$\bf M $}\put(0,-0.4){$ \bf A $ }\put(4,-0.4){$\bf <klux>B</klux> $ }\put(2,4.3){$ \bf C $ }\qbezier(0.5,0)(0.4,0.9)(0.3,0.65)\qbezier(1.8,3.5)(2,3)(2.3,3.5)\qbezier(3.5,0)(3.6,0.5)(3.7,0.65)\put(3,2){\line(-3,-2){3}}\put(3.2,1.7){\line(-3,-2){0.36}}\put(2.9,1.45){\line(-1,2){0.18}}\put(3.3,2.3){$ \bf N $}\end{picture}$