Geometry Challenge – 17

Proof: Rotating $\triangle{ADC}$ counter-clock-wise around $C$ by $60^\circ$, so that $D$ is moved to $D’$, and $A$ to $B$, we have:

Because $\angle{DCD’}=60^\circ$, and $CD=CD’$, $\triangle{DCD’}$ is equilateral. Therefore, $CD=DD’$, and $\angle{CDD’}=60^\circ$.

Because $AD=BD’$, therefore the line segment $AD$, $BD$ and $CD$ forms $\triangle{DBD’}$.

Because $\angle{BDC}=150^\circ$, $\angle{CDD’}=60^\circ$, we have $$\angle{BDD’}=\angle{BDC}-\angle{CDD’}=150^\circ-60^\circ=90^\circ$$

Therefore, $\triangle{DBD’}$ is right, implying the line segment $AD$, $BD$ and $CD$ are the sides of a right triangle.