An angle in a regular hexagon

Question

The marked angle is $60^\circ$.

But I wonder how the angle is found without using trigonometry.

EDIT.

Extra question: How can it be proved that the line DI bisects the angle GIH?

Draw all the corresponding segments like the ones shown. ($C$ to the midpoint of $EF$, etc.) You'll get a symmetric figure that makes the calculation clear. — Blue, Sep 30 '18 at 02:50
Alternative clue: the two quadrilaterals $ABCG$ and $BCDH$ are related how? — Lubin, Sep 30 '18 at 02:53
Not your foolishness, only your failure to look at the diagram in just the “right” way. In mathematics, two people look at the same phenomenon and see it entirely differently, and prove entirely different theorems about the phenomenon. — Lubin, Sep 30 '18 at 03:14
Join DI and extend it to J. One more triangle like ABCQ, BCDH is thus made. Then it is easier to see bisection. — Narasimham, Sep 30 '18 at 08:53
@Narasimham Sorry. I couldn’t catch your answer. Could you let me know it in detail? — P.-S. Park, Oct 01 '18 at 13:02

score 0 · Answer 1 · answered Oct 02 '18 at 07:08

0

hexagon[1]

Could be a helpful construction. 60 deg is formed by rotation in symmetry, but DI not showing as bisector.

answered Oct 02 '18 at 07:08

Narasimham

1 Answers1