Question

Prove that an equlaterial triangle is equiangular

(please write some in mathematics order using x don't just explain in words) ​

Answer 1

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 .

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

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

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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

• $\sf \blue{AC = BC}$
• $\sf\blue {\angle ACM = \angle BCM}$
• $\sf\blue {\angle CAM= \angle CBM}$

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$\underline{\textsf{\textbf{\green{\orange{ \mapsto } In\: \triangle \:ACN \:and\: \triangle \:ABN}}}}$

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 2

$\large \bf \mapsto \color{red} question$

Prove that an equlaterial triangle is equiangular.

$\large\mapsto \color{blue}\bf given$

M∠A = M ∠B = M∠C

$\bf \large \mapsto\color{purple} proof$

In Triangle ABC,

AB = AC.......... (Given)

M ∠ACB = M∠ ABC. ....... (1)angle opposite id congruent side.

But AB = BC.........Given

M ∠ACB = M ∠BAC ...... (2)angle opposite of congruent side.

Therefore, for equation (1) and (2)

M ∠ABC = M ∠ACB = M ∠BAC

So, we can say that M∠A = M∠B = M∠C