kominis3q

2022-07-16

In this image, $MN\parallel AT$, $CD=DA$ and $\mathrm{\angle }PDT=90º$ then arc $NP$=?

Ryan Hahn

Expert

Explanation:
Notice that DT is pependicular bisector for segment AC.
Since $\mathrm{\angle }NMT=\mathrm{\angle }MTA=\mathrm{\angle }MTA$ we see that chord CM and NT are equal. So CMTN is trapezoid and so CN||MT and thus $PC\mathrm{\perp }CN$, so $\mathrm{\angle }NCP={90}^{\circ }$

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