Fórum: quadrilateral

The motivation for the previous solution was a property of the diagonals of a regular 18-gon: their angle is always a multiple of \(\displaystyle 10 ^\circ\).

Also, we can choose three of these diagonals in many ways so that the three chosen diagonals are concurrent.

Nice solution.

Another approach could be to draw a \(\displaystyle BCG\) equilateral triangle and prove by contradiction that \(\displaystyle G\) should line on the segment \(\displaystyle AD\).

The bisector of section AC intersects section AD at E, and the bisector of section BD intersects section AD at F (ABCD is a convex quadrilateral).

Due to AB=BC=CD, the ABCE and the BCDF quadrilaterals are deltoids, FBC angle is 5x and EBC angle is also 10x/2=5x. E is the same F because both points lie on the segment AD. Hereinafter, the notation E is used.

Then AEB angle = BEC angle = CED angle = 60, hence 7x + 5x = 12x in the ABE triangle = 120. There is x = 10 and in the convex quadrilateral ABCD the angle at C is 360-22x = 140.

The triangles ABE, BEC, and CED are congruent, with angles of 50, 60, and 70, respectively.

yes but how

140, x=10

Given a convex quadrilateral \(\displaystyle ABCD \) with \(\displaystyle AB = BC = CD \) and \(\displaystyle m(\angle DAB) = {7x }\),\(\displaystyle \,m(\angle A BC) = {10x }\), \(\displaystyle m(\angle CDA) = { 5x }\), Find \(\displaystyle m(\angle BCD) \).