If point o moves to point b on AB, with point o as the center,
The condition that the circle with radius OB still intersects BC and D and DE⊥AC remains unchanged,
The above conclusion holds.
The reasons are as follows: even OD.
Evidence 1:
AB = AC
∴∠B = ∠C - ①
OB = OD
∴∠B = ∠ODB - ②
From ① ②: ∠ C = ∠ ODB-③
∴ OD ‖ AC
You ∵ DE⊥AC
∴ de⊥ ·⊥od and d points are on the circle o,
∴DE is the tangent of circle O. (The tangent of the circle is perpendicular to the radius of the tangent point)
Evidence 2:
AB = AC
∴∠B = ∠C - ①
OB = OD
∴∠B = ∠ODB - ②
From ① ②: ∠ C = ∠ ODB-③
∵DE⊥AC
∴∠C + ∠EDC = 90 - ④
From ③ ④: ∠ ODB+∠ EDC = 90.
∴∠ODE = 180 - (∠ODB + ∠EDC)
= 180 - 90 = 90
∴ de ⊥ OD. Point D is on the circle O,
∴DE is the tangent of circle O. (The tangent of the circle is perpendicular to the radius of the tangent point)
Second question (2) When the center of the circle O moves to a position far away from point B 15/8 on AB,
Circle o is tangent to AC. The reason for this is the following：
Solution 1:
Let circle o be tangent to AC at point m,
Even OM, and OM ⊥ AC (the tangent of a circle is perpendicular to the radius of the tangent point).
Let the circle O be tangent to AC, and the radius of the circle O is ob = om = r.
Then AO = AB-OB = (5-r)
At Rt△AOF.
∵ sinA = OM / AO
That is, 3/5 = r/(5-r)
∴ 5r = 3(5 - r)
∴ r = 15/8
That is, when the center of O moves to the position 15/8 on AB far from point B,
Circle o is tangent to AC.
Solution 2: Let circles O and AC be tangent to point M,
Even OM, and OM ⊥ AC (the tangent of a circle is perpendicular to the radius of the tangent point).
Let the circle O be tangent to AC, and the radius of the circle O is ob = om = r.
Then AO = AB-OB = (5-r)
If b is BN ⊥ AC in n, then OM ‖ BN.
In Rt△ABN, BN = AB×sinA = 5×(3/5)= 3.
From OM BN:
OM :BN = AO :AB
That is, r: 3 = (5-r): 5.
∴ 5r = 3(5 - r)
∴ r = 15/8
That is, when the center o moves to the position 15/8 on AB far from point B,
Circle o is tangent to AC.
Solution 3: Let circles O and AC be tangent to point M,
Even OM, and OM ⊥ AC (the tangent of a circle is perpendicular to the radius of the tangent point).
Let the circle O be tangent to AC, and the radius of the circle O is ob = om = r.
Then AO = AB-OB = (5-r)
Let c be CH ⊥ AB in h (solved by the area equation of △AOC)
Then in Rt△ACH, CH = AC×sinA = 5×(3/5)= 3.
∫S△AOC =( 1/2)×AO×CH =( 1/2)×OM×AC
∴ AO × CH = OM × AC
∴(5 - r)× 3 = r × 5
∴ r = 15/8
That is, when the center o moves to the position 15/8 on AB far from point B,
Circle o is tangent to AC.
Please pay attention to multiple solutions to one question, broaden your thinking and cultivate your habit of in-depth inquiry.
Some solutions to this problem are a bit cumbersome, but they are more concise when solving other problems.
In addition, please pay attention to the variable and extension of the topic.
① If the acute angle ∠BAC in this question is replaced by the obtuse angle ∠BAC, the result is the same.
② If the first question is changed to "If point O on AB moves to point A,
Make a circle with point O as the center and OA as the radius, and intersect AC at E ".
So there is a new conclusion: BE is the tangent of circle O.
③ If the first question is changed to "If AB=AC=5 cm, sinA=3/5"
If point O moves to point A on AB, make a circle with point O as the center and OA as the radius.
Find the point O to move to the position of AB, and the circle O is tangent to BC ",
This problem is a bit difficult, and the result is OA = 15 (√ 10-3).
For the senior high school entrance examination, don't do too many questions, but experience and summarize the typical questions.
It will be of great benefit to senior high school to draw inferences from others, dig deeper and cultivate their own sense of inquiry and innovation.