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Anonymous users2024-02-151) Solution: ab cd, acd+ cab=18o°, acd=114°, cab=66°, from the practice, am is the bisector of acb, amb=one-half cab=33°
2) Proof of: am bisect cab, cam= mab, ab cd, mab= cma, cam= cma, cn am, anc= mnc, in acn and mcn, anc= mnc , cam= mac , cn=cn , acn mcn.
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Anonymous users2024-02-14Register and log in on the test question page on the home page.
Reveal the truth behind the test questions Question answers, test points, combing, and drawing inferences.
Question answer. Inscription.
As shown in the figure, AB CD, with point A as the center of the circle, less than the length of AC as the radius as the arc, respectively intersect AB, AC at E, F two points, and then respectively with E, F as the center of the circle, greater than EF length as the radius as the arc, the two arcs intersect at the point P, as the ray AP, and the intersection of CD at the point M.
1) If acd=114°, find the degree of mAb;
2) If cn am and the vertical foot is n, verify that it is acn mcn.
Answer (Find homework answers --- on the Rubik's Cube).
1) Solution: ab cd, acd+ cab=18o°, acd=114°, cab=66°, from the practice, am is the bisector of acb, amb= cab=33°
2) Proof of: am bisect cab, cam= mab, ab cd, mab= cma, cam= cma, cn am, anc= mnc, in acn and mcn, anc= mnc , cam= mac , cn=cn , acn mcn.
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Anonymous users2024-02-13Thank you! I'll help you write the answer
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Anonymous users2024-02-12Give the answer by giving the evaluation first.
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Anonymous users2024-02-11Let the diameter be x
So the circumference of the circle is x
Columnable equations. 2x＋∏x＝
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Anonymous users2024-02-10Connecting PA and QA, because PM is vertically bisected AB, QN is vertically bisected AC, so ABP and Xinqinglu AQC are isosceles triangles, so PBA is equal to PAB, QAC is equal to slippery QCA, so Paq BAC ( BAP QAC) is mu BAC ( CBA BCA) BAC (180° BAC) 30°
paq＝30°
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