Point a and B are two points on O, ab = 10, point P is a moving point on O (P does not coincide with ab) to connect AP, Pb passes through point o as OE ⊥ AP, of ⊥ Pb, and point EF is (1) Find the distance from the long (2) point O of the line EF to AB, and find the radius of O

Point a and B are two points on O, ab = 10, point P is a moving point on O (P does not coincide with ab) to connect AP, Pb passes through point o as OE ⊥ AP, of ⊥ Pb, and point EF is (1) Find the distance from the long (2) point O of the line EF to AB, and find the radius of O

1. EF is the midpoint of PA and Pb sides of △ PAB
EF=AB/2=5
2、R^2=2^2+(AB/2)^2=4+25=29
R = root 29

As shown in the figure, in the triangle ABC, the bisector AP of the angle BAC and the vertical bisector PD of the BC side intersect at point P, make PM, make AB perpendicular to point m, make PN perpendicular to AC, and make the extension line of AC perpendicular to point n. please specify BM = cn

Question supplement: make ad vertical BC through a, connect a point on AC, and turn ad to P. there's no graph. I also said a bunch of E? P is the midpoint of AD, connect BP and extend BP, and turn AC to e BC = (√ 5) AB AC: BC=

Take any two numbers P and Q (P ≠ q) from the four numbers 2, 3, 4 and 5 to form the functions Y1 = px-2 and y2 = x + Q, so that the intersection of the two function images is on the left side of the straight line x = 2, then such ordered array (P, q) has ()
A. 4 groups B. 5 groups C. 6 groups D. uncertain

Let px-2 = x + Q, the solution is x = q + 2p − 1, because the intersection point is on the left side of the line x = 2, that is, q + 2p − 1 ＜ 2, then q ＜ 2p-4 is sorted out. By substituting P = 2, 3, 4, 5 respectively, the corresponding value of Q can be obtained. The ordinal number pairs are (4, 2), (4, 3), (5, 2), (5, 3), (5, 4), (5, 5), and because P ≠ Q, so (5, 5) is rounded off, and there are 5 pairs that satisfy the condition