In this image, MN∥AT, <math xmlns="http://www.w3.org/1998/Math/MathML"> <mi>C</mi> <mi>D</mi> <mo>=</mo> <mi>D</mi> <mi>A</mi> </math> and <math xmlns="http://www.w3.org/1998/Math/MathML"> <mi mathvariant="normal">&#x2220;<!-- ∠ --></mi> <mi>P</mi> <mi>D</mi> <mi>T</mi> <mo>=</mo> <mn>90</mn> <mrow class="MJX-TeXAtom-ORD"> <mo>&#xBA;</mo> </mrow> </math> then arc <math xmlns="http://www.w3.org/1998/Math/MathML"> <mi>N</mi> <mi>P</mi> <mo>=</mo> <mo>?</mo> </math>

kominis3q

kominis3q

Answered question

2022-07-16

In this image, M N A T, C D = D A and P D T = 90 º then arc N P=?

Answer & Explanation

Ryan Hahn

Ryan Hahn

Beginner2022-07-17Added 11 answers

Explanation:
Notice that DT is pependicular bisector for segment AC.
Since N M T = M T A = M T A we see that chord CM and NT are equal. So CMTN is trapezoid and so CN||MT and thus P C C N, so N C P = 90

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