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In a neutral geometry, given $\triangle{ABC}$ with $A-D-B,$ $\ A-E-C,$ $\angle{ABE}\cong \angle{ACD},$ $\angle{BDC}\cong \angle{BEC}$ and the line segment $\overline{BE}\cong \overline{CD}$, then $\triangle{ABC}$ is an isosceles.

$A-D-B$ means the point $D$ is between $A$ and $B$.

Since we are in a neutral geometry then our traigle satisfies the side angle side axiom.

How can I prove this?

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$\triangle ADC\cong\triangle AEB$ by ASA:

  • A: $\angle ABE=\angle ACD$
  • S: $BE=CD$
  • A: $\angle ADC=\angle ADB-\angle BDC=\angle AEC-\angle BEC=\angle AEB$

So $AC\cong AB$ as corresponding sides.

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  • $\begingroup$ How did you get that $\triangle ADC\cong\triangle AEB $? $\endgroup$ – Alex Apr 17 '14 at 2:03
  • $\begingroup$ @Alex : I expanded that part of the argument. $\endgroup$ – user21467 Apr 17 '14 at 10:46

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