Was there a typo? I'm pretty sure the rectangle would be a square if AB = AD

Also, ABC would be a right angle

either my translation using google translate is not ok, or there is a typo, here is the wording and the source's photo ''Diberikan segi empat talibusur ABCD sehingga AB = AD dan AB+BC < CD. Buktikan bahwa sudut ABC lebih dari 120 derajat '' Indonesian is the language

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Maybe it is just a quadrilateral and not a rectangle? Just guessing, as I don't know this language @below thanks

It was not meant to be a rectangle; just a cyclic quadrilateral. Is that a google translation? (I know Google translate sucks but what...) "Given a cyclic quadrilateral ABCD such that $AB = AD$ and $AB + BC < CD$. Prove that the angle $ABC$ is more than $120$ degrees."

Hint

BlazingMuddy wrote: It was not meant to be a rectangle; just a cyclic quadrilateral. Is that a google translation? (I know Google translate sucks but what...) "Given a cyclic quadrilateral ABCD such that $AB = AD$ and $AB + BC < CD$. Prove that the angle $ABC$ is more than $120$ degrees."

Hint

can you explain your solution?

BlazingMuddy wrote: It was not meant to be a rectangle; just a cyclic quadrilateral. Is that a google translation? (I know Google translate sucks but what...) "Given a cyclic quadrilateral ABCD such that $AB = AD$ and $AB + BC < CD$. Prove that the angle $ABC$ is more than $120$ degrees."

Hint

Can you explain more about your solution? My try is this, $AB=AD$ imply that $AC$ is angle bisector of $\angle BCD$, Also, if we write $x=BD, AB=AD=s,BC=t, CD=u$ we get $s+t<u$ and $x\left(s^2+t^2-2st\cos \angle ABC\right) > s^2+st.$ Then I'm stuck.

let $r= \frac{1}{2}$ let $ADB = a$ $DAC=b$ $AB=AD = \sin(a)=x$ $AC=\sin(a+b)=x*\sqrt{1-y^2}+y*\sqrt{1-x^2}$ $DC = \sin(b)=y$ $BD = \sin(2a)=2x*\sqrt{1-x^2}$ Ptolemy $AB*DC+BC*AD = BD*AC$ $x*(y+BC) = 2x*\sqrt{1-x^2}*(x*\sqrt{1-y^2}+y\sqrt{1-x^2})$ $BC =2x\sqrt{1-x^2}*\sqrt{1-y^2}+y-2x^2y$ $x+y-2x^2y+2x\sqrt{1-x^2}*\sqrt{1-y^2}<y$ $2xy-1>2\sqrt{1-x^2}\sqrt{1-y^2}$ $4y^2+4x^2-4xy >3$ $(x\sqrt{1-y^2}+y\sqrt{1-x^2}) ^2 = x^2+y^2-2xy*\cos(180-a-b)$ $x^2+y^2-2x^2y^2+2xy*\sqrt{1-x^2}*\sqrt{1-y^2}=x^2+y^2-2xy*cos(180-a-b)$ $-2x^2y^2+2xy\sqrt{1-x^2}\sqrt{1-y^2} = -2xy*cos(180-a-b)$ $\frac{1}{2}<xy-\sqrt{1-x^2}\sqrt{1-y^2} = cos(180-a-b)=-cos(a+b)$ $cos(a+b)<-\frac{1}{2}$ $cos(a+b)<cos(120)$ $120 <a+b$