# Prove that in quadrilateral $ABCD$, $|\angle DAB|=|\angle ABC|=|\angle BCD|$, $|DA|=|DC|$, if $|BE|=|DA|$, $CE$ is the angle bisector of $\angle BCD$

When trying to solve this problem in the usual way - playing around with angles in the quadrilateral - I always seemed to be lacking one piece of information. I am certain there is something crucial to be extracted out of the fact, that three of the angles and two of the sides are equal. I have never however come across such an object (nor do I know if it even has a name or whether it bears some crucial reseamblance to the parallelogram in terms of properties) so I wondered which useful or peculiar properties might be of interest here.

• Where does $E$ come from? Commented Jun 24, 2021 at 22:52
• @GoRza i believe E lies on AD and is the point of intersection of the bisector of BCD and AD Commented Jun 24, 2021 at 23:06
• I think you could show that BA=BC.
– Moti
Commented Jun 25, 2021 at 4:22

First a small note: please post figures in geometry questions. It helps readers better understand your question and your work. In the case of this question there was missing information in the text about point $$E$$ and I follow the assumption of user C Squared, i.e. that $$E$$ is on the segment $$AB$$. If this assumption is wrong, please let me know.

In the above figure, the equal-length segments $$AD$$, $$CD$$ and $$BE$$ are shown in red. Blue segments are added to the figure as I shall describe. Alright, here we go:

$$\triangle ACD$$ is isosceles, therefore $$\widehat{CAD}=\widehat{ACD}$$

This means that $$\triangle ACB$$ is also isosceles, because $$\widehat{CAB} = \widehat{DAB} - \widehat{CAD} = \widehat{DCB} - \widehat{ACD} = \widehat{ACB}$$ It follows that the segment drawn from $$D$$ to the midpoint of $$AC$$ is the perpendicular bisector of $$AC$$, and it is also the bisector of $$\widehat{ADC}$$. Similarly, the segment drawn from $$B$$ to the midpoint of $$AC$$ is the perpendicular bisector of $$AC$$ and also the bisector of $$\widehat{ABC}$$. Note that because $$AC$$ has only one perpendicular bisector, $$BD$$ is the perpendicular bisector of $$AC$$, and at the same time $$BD$$ bisects $$\widehat{ABC}$$ and $$\widehat{ADC}$$ .

Now let's draw the segment $$DE$$ . In the quadrilateral $$BCDE$$ , $$\widehat{EBC}=\widehat{DCB}$$ and $$BE=CD$$ . The interested reader can use this information to prove that this quadrilateral is an isosceles trapezoid. It follows that the four points $$B$$, $$C$$, $$D$$ and $$E$$ are on the same circle.

Can you take it from here?

• I feel that you could have left the last 2 to 3 steps because helping OP with full solution isn't wrong but still should be avoided. Try to provide OP with some tricks and write at end something like Can you continue it OP? If OP would have problem he would write in comments and then you can help him by writing in comment or editing the question.
– user876009
Commented Jun 25, 2021 at 5:22
• @JitendraSingh You are right. I will edit my post, just in case OP hasn't seen it yet. Commented Jun 25, 2021 at 5:24
• Well that would be good enough. Have a upvote from me
– user876009
Commented Jun 25, 2021 at 5:25
• @JitendraSingh Thank you! Commented Jun 25, 2021 at 5:28

The question is ambiguous about where point $$E$$ lies but assuming that it lies on $$AB$$ satisfies all the conditions so I am assuming that is where it is supposed to be.

Observe that, $$ABCD$$ is a kite with $$DA=DC$$ and $$AB=BC$$. Now, it can be proven by simple angle chasing that $$DEBC$$ is an isosceles trapezoid and $$DE=DC$$.

Hence, $$\angle DCE=\angle DEC=\angle BCE$$ and therefore $$CE$$ is the bisector of $$\angle BCD$$.