Question 388334
You've just shown that the midpoint of AC is *[Tex \LARGE \left(\frac{x_{2}}{2} \ ,\frac{y_{2}}{2}\right)]



Now find the midpoint of BD to get *[Tex \LARGE \left(\frac{x_{1}+x_{3}}{2} \ ,\frac{y_{3}}{2}\right)]



Since the AC and BD have the same midpoint, this means that *[Tex \LARGE \left(\frac{x_{2}}{2} \ ,\frac{y_{2}}{2}\right)=\left(\frac{x_{1}+x_{3}}{2} \ ,\frac{y_{3}}{2}\right)]



Which therefore means that *[Tex \LARGE \frac{x_{2}}{2}=\frac{x_{1}+x_{3}}{2} \ ,\frac{y_{2}}{2}=\frac{y_{3}}{2}]



I'll leave the rest to you. Let me know if you need more help.



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