Question 252352
In ΔABC, Given (a/cosA)=(b/cosB)=(c/cosC) . Prove that ΔABC is an equilateral triangle.

{{{a/(cosA)=b/(cosB)=c/(cosC)}}}
<pre><font size = 4 color = "indigo"><b>
By the law of sines:

{{{a/(sinA)=b/(sinB)=c/(sinC)}}}

and therefore

{{{(sinA)/a=(sinB)/b=(sinC)/c}}}

Multiplying equals by equals:

{{{((sinA)/a)(a/(cosA))=((sinB)/b)(b/(cosB))=((sinC)/c)(c/(cosC))}}}

{{{((sinA)/cross(a))(cross(a)/(cosA))=((sinB)/cross(b))(cross(b)/(cosB))=((sinC)/cross(c))(cross(c)/(cosC))}}}

{{{((sinA)/(cosA))=((sinB)/(cosB))=((sinC)/(cosC))}}}

{{{tanA=tanB=tanC}}}

A,B, and C are angles of a triangle and thus are less than 180°.

Any two angles less than 180° which have the same tangent must
be equal in measure.  Thus

A = B = C

Thus triangle ABC is equiangular and every equiangular
triangle is also equilateral.

Edwin</pre>