Take P on the left side of the arc as an example:
It is proved that △BHK is similar to △BGM and △BHK is similar to △HAO, and then BK= 1/3 is obtained by using some properties of the ratio.
The specific proof is as follows:
Solution: ∫ square ABCD, side length is 2, O is the midpoint of AD.
∴AO= 1,∠ABC=90
And ∵O is the center of the circle, OE is the radius, and the straight line MPG is the tangent of the circle O.
∴oh⊥mg ∴op⊥mpg
∴∠MPH=90 ∴∠BHP+∠BMG=90
In the right triangle BMG, ∠ bgm+∠ BMG = 90.
∴∠BHP=∠BGM
And because: ∠ HBK = ∠ GBM = 90 = ∠ A.
∴△BHK is similar to△ △BGM, and△ bhk is similar to△ Hao.
∴BG/BH=BM/BK,BK/AO=BH/AH
∴BG/BM=BH/BK=3,BH/BK=AH/AO=3
∴BH= 1,BK= 1/3
In the second case, when the point P is on the right side of the arc, it is similar to a triangle, and the corresponding sides are proportional. Then we can get it by using some properties of the ratio:
BK=5/3
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