Given a hexagon ABCDEF inscribed in a circle with AB=BC, CD=DE, EF=FA, show that AD, BE and CF
are concurrent.
+397 This went through a few weeks ago didn't it ?
I take it that no solution was posted back then, in which case, here's my solution.
This is best done on paper, filling in angles as we go.
Start by joining each point to every other point, so in addition to what you see in the diagram above, join A to C , A to E, B to D, B to F, etc..
Beginning with the triangle ABC, it's isosceles since AB = BC, in which case angle ACB = angle BAC.
Let angle ACB = 'circle', (draw a little circle on the diagram),
then, using the angles on the same arc property, all other angles on the arc AB will equal 'circle',
ie. angle ACB = angle ADB = angle AEB = angle AFB = 'circle',
and since angle ACB = angle BAC,
angle BAC = angle BDC = angle BEC = angle BFC = 'circle'.
Moving on to the triangle CDE, and using the same reasoning as above, let angle CED = 'cross', then
angle CED = angle CAD = angle CBD = angle CFD = 'cross',
angle DCE = angle DAE = angle DBE = angle DFE = 'cross'.
Onto triangle AEF, let angle AEF = 'dot', then
angle AEF = angle ABF = angle ACF = angle ADF = 'dot',
angle EAF = angle EBF = angle ECF = angle EDF = 'dot'.
(Not all of those angles are necessary for the solution, but it just seemed easier to keep going.)
Now look at the triangle ACE, it has two crosses at A, two dots at C and two circles at E.
That means that 2 crosses + 2 dots + 2 circles = 180 deg, so 'cross' + 'dot' + 'circle'= 90 deg.
Next, look at the intersection of BD and FC, call that point G and consider the triangle GCD.
Angle C = 'dot' + 'cross' and angle D = 'circle', so angle GCD + angle GDC = 'dot' + 'cross' + 'circle' = 90 deg,
in which case, angle DGC = 90deg, so FC is perpendicular to BD.
Now consider the triangle, BDF, from the last line above, it follows that FG (extended to C) is an altitude of this triangle.
Similarly BE and DA are also altitudes of the triangle BDF and these are known to be concurrent, hence FC, BE and DA are concurrent.
+397 This went through a few weeks ago didn't it ?
I take it that no solution was posted back then, in which case, here's my solution.
This is best done on paper, filling in angles as we go.
Start by joining each point to every other point, so in addition to what you see in the diagram above, join A to C , A to E, B to D, B to F, etc..
Beginning with the triangle ABC, it's isosceles since AB = BC, in which case angle ACB = angle BAC.
Let angle ACB = 'circle', (draw a little circle on the diagram),
then, using the angles on the same arc property, all other angles on the arc AB will equal 'circle',
ie. angle ACB = angle ADB = angle AEB = angle AFB = 'circle',
and since angle ACB = angle BAC,
angle BAC = angle BDC = angle BEC = angle BFC = 'circle'.
Moving on to the triangle CDE, and using the same reasoning as above, let angle CED = 'cross', then
angle CED = angle CAD = angle CBD = angle CFD = 'cross',
angle DCE = angle DAE = angle DBE = angle DFE = 'cross'.
Onto triangle AEF, let angle AEF = 'dot', then
angle AEF = angle ABF = angle ACF = angle ADF = 'dot',
angle EAF = angle EBF = angle ECF = angle EDF = 'dot'.
(Not all of those angles are necessary for the solution, but it just seemed easier to keep going.)
Now look at the triangle ACE, it has two crosses at A, two dots at C and two circles at E.
That means that 2 crosses + 2 dots + 2 circles = 180 deg, so 'cross' + 'dot' + 'circle'= 90 deg.
Next, look at the intersection of BD and FC, call that point G and consider the triangle GCD.
Angle C = 'dot' + 'cross' and angle D = 'circle', so angle GCD + angle GDC = 'dot' + 'cross' + 'circle' = 90 deg,
in which case, angle DGC = 90deg, so FC is perpendicular to BD.
Now consider the triangle, BDF, from the last line above, it follows that FG (extended to C) is an altitude of this triangle.
Similarly BE and DA are also altitudes of the triangle BDF and these are known to be concurrent, hence FC, BE and DA are concurrent.
