Let $\angle CAB=\alpha$ and $\angle DAC=\beta$. Since $AN=AB=AD$ then 1)$\angle DNA=\frac{\pi-\angle DAN}{2}$ and $\angle DAN=\angle CAN-\beta=\angle CAB-\beta=\alpha-\beta$ 2)$\angle AEB=\pi-2\alpha$ 3)$\angle ADC= \angle DCB=\angle DCA + \angle ACB=2\alpha$ thus in triangle $ADC$ $\pi=3\alpha+\beta$ From 1), 2) and 3) we get $\angle AND+\angle AED =\pi$

parmenides51 wrote: Let $ABCD$ be an isosceles trapezoid with $AB=AD=BC, AB//CD, AB>CD$. Let $E= AC \cap BD$ and $N$ symmetric to $B$ wrt $AC$. Prove that the quadrilateral $ANDE$ is cyclic. Does $//$ mean parallel?

programjames1 wrote: ... Does $//$ mean parallel? yes

Let \( ABCD \) be an isosceles trapezoid, where \( AB \parallel CD \) and \( ABCD \) is cyclic. Given that \( \angle CAB = \angle ACB = a \), we can deduce the following: Since \( AB \parallel CD \), it follows that \( \angle CAB = \angle DCA = a \). Also, because \( ABCD \) is cyclic, we have \( \angle DCA = \angle DBA = a \). Let \( N \) be a point such that \( NC \parallel AB \), then \( \angle NDB = 180^\circ - a \). Let \( H \) be the intersection point of \( AC \) and \( NB \). Since \( N \) is symmetric to \( B \), we have \( NH = HB \) and \( AH = HC \). Since triangle \( ABC \) is isosceles and \( AH = HC \), it follows that \( \angle CHB = 90^\circ \). Additionally, since \( \angle AHB = 90^\circ \) and \( NH = HB \), we have that \( AH \) is the angle bisector, and thus \( \angle HAB = \angle HAN = a \). Finally, we have \( \angle NAE + \angle NDE = a + (180^\circ - a) = 180^\circ \), which implies that quadrilateral \( ANDE \) is cyclic.

see the figure

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