Basic Trigonometry Proof

[Vertciel]

Hello everyone,

Here is my problem:

In a triangle ABC, we know that b(cosA) = a(cosB). Prove that the triangle is isoceles. (Hint: Use The Cosine Law)

I've tried to isolate b, a, cos A, and cosB from b(cosA) = a(cosB) and substitute them into the law of cosines. This has gotten me nowhere though.

Could anyone please just give a hint?

Thanks.

 

[soroban]

Hello, Vertciel!

n $\displaystyle \Delta ABC$, we know that: $\displaystyle \:b\cdot\cos A \:= \:a\cdot\cos B$
Prove that the triangle is isosceles. .(Hint: Use The Cosine Law)

I haven't found a way to apply the Cosine Law (yet) . . .

            C
            *
           *| *
          * |   *  a
       b *  |     * 
        *   |       *
       *    |         *
    A * * * + * * * * * * B
            D

Draw altitude $\displaystyle CD$ from vertex $\displaystyle C$ to side $\displaystyle AB$.

In right triangle $\displaystyle CDA:\;\cos A \:= \:\frac{AD}{b}\;\;\Rightarrow\;\;AD \:=\:b\cdot\cos A$

In right triangle $\displaystyle CDB:\;\cos B \:=\:\frac{DB}{a}\;\;\Rightarrow\;\;DB \:=\:a\cdot\cos B$

Since $\displaystyle \,b\cdot\cos A \:=\:a\cdot\cos B$, then: \(\displaystyle \,AD\:=\B\)


In right triangle $\displaystyle CDA$ and $\displaystyle CDB$, we have: $\displaystyle \:\begin{Bmatrix}AD & = & DB \\ \angle CDA & = & \angle CDB \\ CD & = & CD\end{Bmatrix}$

Hence: $\displaystyle \:\Delta CDA$ is congruent to $\displaystyle \Delta CBD\;$ (s.a.s)

Therefore: $\displaystyle AC \,=\,BC$ (corresponding parts)

. . and $\displaystyle \Delta ABC\text{ is isosceles.}$

 

[pka]

\(\displaystyle \L\begin{array}{rcrr}
\left. {\begin{array}{r}
{a^2 = b^2 + c^2 - 2bc\cos (A)} \\
{b^2 = a^2 + c^2 - 2ac\cos (B)} \\
\end{array}} \right\} &\Rightarrow & \quad \frac{{b^2 + c^2 - a^2 }}{{2c}}& =& \frac{{a^2 + c^2 - b^2 }}{{2c}} \\
&\Rightarrow &2b^2 + c^2 &=& 2a^2 + c^2 \\
&\Rightarrow& b& =& a \\
\end{array}\)

 

[Vertciel]

\(\displaystyle \L\begin{array}{rcrr}
\left. {\begin{array}{r}
{a^2 = b^2 + c^2 - 2bc\cos (A)} \\
{b^2 = a^2 + c^2 - 2ac\cos (B)} \\
\end{array}} \right\} &\Rightarrow & \quad \frac{{b^2 + c^2 - a^2 }}{{2c}}& =& \frac{{a^2 + c^2 - b^2 }}{{2c}} \\\)

Thank you soroban and pka for your assistance.

I have a question for pka's solution:

Shouldn't cos A = (b^2 + c^2 - a^2)/2bc and cos B = (a^2 + c^2 - b^2)/2ac? If so, how did you eliminate the a and b in the denominators? Did you eliminate the a and b on the basis that a = b?

Thanks again.

 

[tkhunny]

Look VERY carefully at your given information.

Do NOT solve for the cosine alone.

 

[pka]

\(\displaystyle \L b\cos (A) = \frac{{b^2 + c^2 - a^2 }}{{2c}}\)

 

[Vertciel]

Thank you very much for your assistance and your reply.