Isosceles triangles and deltoids.

Question

An ABC triangle has an AD edge that comes over the BC edge over angle A. The ABD angle is 20 degrees and the ACD angle is 40 degrees. Find the degree of BAD angle

Answer 1

Extend $BC$ and make $CE=BD$. Then $\triangle{ACE}$ is isosceles.

$\angle{ACE}=140^{\circ}$ so $\angle{CAE}=\angle{CEA}=20^{\circ}$. Now we see that $\triangle{ABE}$ is isosceles and $AE=AB$. Furthermore, $\triangle{ADE} \cong \triangle{ABC}$ therefore $\angle{ADE}=\angle{ACB}=40^{\circ}$. Finally, $x=20^{\circ}$ because $AD=AC=BD$.

Answer 2

Use the sine law: $$ \sin20°:AD=\sin x:BD,\quad \sin40°:AD=\sin(20°+x):AC. $$ Dividing the second equation by the first yields: $$ \sin 40°:\sin20°=\sin(20°+x):\sin x, $$ an equation easy to solve for $x$.