I have been working on the following tasks for some time now and I do not know how to solve it.
$(A, B, C)$ is a triangle and $K$ is his nine point circle. Show it:
a) $(A, B, C)$ is right-angled if and only if one of the points $A, B, C$ lies on $K$.
b) If the line $L (A, B)$ is tangent to $K$ then $a = b$
Hope someone can help me. Thanks
Since it seems that the problem is related to math olympiad/contest, I will write the answer in that sense.
Let $H$ be the orthocenter of $(A,B,C)$, and let $H_a,H_b,H_c$ be the points on $\overline{BC},\overline{CA},\overline{AB}$, resp, that $\overline{BC}\perp\overline{AH_a},\overline{CA}\perp\overline{BH_b},\overline{AB}\perp\overline{CH_c}$.
a) If $(A,B,C)$ is right-angled, either $H_a=A$, $H_b=B$ or $H_c=C$ should hold and that should be on $K$.
Conversely, assume without loss of generality that $A\in K$, then $A,H_b,H_c,H$ are concyclic(lie on one common circle). Since $A,H_b,H_c,H$ are also concyclic, $A,H_b,H_c,H,H_a$ are concyclic and here $A,H,H_a$ are collinear(line on one common line), two of them should be the same. However, $A\neq H_a$ by definition. Now, $A$ is right angle if $A=H$, and $B$ or $C$ is right angle if $H_a=H$.
b) Since $(H_a,H_b,H_c)$ is tangent to $\overleftrightarrow{AB}$, $\angle AH_cH_b=\angle H_cH_aH_b$. Again, $\angle AH_cH_b=\angle AHHb=\angle BHH_a=\angle BH_cH_a$ and by tangency $\angle BH_cH_a=\angle H_cH_bH_a$.
Thus, $\angle BH_cH_a=\angle H_cH_aH_b\Rightarrow\overleftrightarrow{AB}||\overleftrightarrow{H_aH_b}\Rightarrow\angle ABC=\angle H_bH_aC=\angle BAC\Rightarrow a=b$.
*To understand pf(b), you should note that $AH_bHH_c,BH_aHH_c,AH_bH_aB$ are concyclic squares.