Question

In an isosceles triangle with base a and a lateral side b the vertex angle is equal to ${20}^{\circ }$. Prove that $|a{|}^3+|b{|}^3=3|a||b{|}^2$.

Answer 1

$AC=a$ $AB=BC=b$

Let $M\sum BC$ such that $AM=a$

Hence $\angle BAM={80}^0-{20}^0={60}^0$

Now let $N\sum AB$ such that $AN=a$

Here $AN=AM=MN=a$ and $\angle NMB={180}^0-{80}^0-{60}^0={40}^0$

Now let $k\sum BC$ such that $Nk=a$

Then, $\angle NkM={40}^0$ which says that

$\angle BNk={20}^0$ and from here $Dk=a$

Now $kLD$ altitude of $\Delta BNk$ and $MFD$ altitude of $\Delta AMC$

Here since $BN=b-a$ and $\Delta BLk\cong \Delta AFM$

we obtain $AF=BC=\frac{b-a}{2}$

and $FC=a-\frac{b-a}{2}=\frac{3a-b}{2}$

Also we see that $\Delta ABC\sim \Delta ACM$ which gives

$\frac{MC}{a}=\frac{b}{a}$ or $MC=\frac{b^2}{a}$

Since ${AM}^2-{AF}^2={MC}^2-{FC}^2$ we obtain

$a^3+b^3=3ab$