+0

0
91
3
+19

It seems to be the case that if ABCD is a quadrilateral, $$AB=BC=CD$$ , and $$\angle ABC+\angle BCD = 240^{\circ}$$, then the angle bisectors of $$\angle ABC$$ and $$\angle BCD$$ intersect on $$\overline{AD}$$. Can someone prove or disprove this?

(Note: I reposted this since I left out an important detail previously)

Dec 19, 2018

#1
+94275
+2

I think this may prove it :

Let the sum of angles ABC and BCD = 240°

And let the bisectors of angles ABC and BCD  be as shown

Since AB = BC = CD

Then by SAS , triangle ABE  is congruent to triangle CBE

Similarly, by SAS, triangle CBE is congruent to triangle CDE

So....triangle CDE is congruent to triangle ABE is congruent to triangle CBE

So  angles DEC, AEB and BEC are congruent

But angles ABC and BCD sum to 240

So   1/2 ABC + 1/2 BCD = 120

But 1/2ABC = EBC

And 1/2 of BCD = BCE

So  angles EBC and BCE = 120

So angle BEC = 180 - (m EBC + BCE) = 180 - 120 = 60

But BEC = AEB = DEC.....so....

60 + 60 + 60  =  180

Therefore angle AED must be  straight a straight angle because  angle AEB + angle BEC + angle DEC  =  60 + 60 + 60  = 180

Thus....E must lie on AD.......and E is the intersection of the bisectors of angles ABC and BCD

Dec 19, 2018
edited by CPhill  Dec 19, 2018
#2
+1

Thank you so much!

Guest Dec 19, 2018
#3
+27310
+1

I'm not so sure about this Chris.  In statement: "Then by SAS , triangle ABE  is congruent to triangle CBE" you have assumed that the bisector of ABC meets the bisector of BCD at the same point on AD (and you then proceed to show everything is consistent if this is the case).  In other words, you have assumed what you set out to prove!  I think you should start with the assumption that the two bisectors meet at potentially different points, say E and F, on AD, and then show that E and F must be the same point.

Alan  Dec 19, 2018