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As shown in the figure, in RT △ ABC, ∠ ACB = 90 ° AB = 6, P is the point on AB, connect PC, set ∠ BCP = m ∠ ACP, when AP = 3 / 2, whether there is a positive integer m Make PC perpendicular to ab? If it exists, calculate the value of M. if it does not exist, explain the reason
Suppose there is a positive integer m such that PC is perpendicular to ab. then ∠ APC = 90 °
Because ∠ BCP + ∠ ACP = ∠ ACB = 90 ° and ∠ ACP + ∠ cab = 90 °
Therefore, Rb △ ABC is similar to RT △ ACP
Similarly, ∠ CBA = ∠ ACP, so ∠ cab = m ∠ CBA, M = ∠ cab / ∠ CBA
From the similarity between RT △ ABC and RT △ ACP, it is concluded that
AC / AP = AB / AC, that is, AC / (3 / 2) = 6 / AC, AC = 3
Sin（∠CBA）=AC/AB=3/6=1/2
Therefore, cab = 60 degree
m=∠CAB/∠CBA=60/30=2
So m exists and the hypothesis holds
As shown in the figure, if △ ACP ∽ ABC, AP = 4, ab = 5, then AC
The corresponding edge of a similar triangle is proportional to AC: ab = AP: AC, so AC is equal to 20 (the square root of 20) under the root sign
The square of AC = AP times AB, proving that ACP is similar to ABC
AC²=AP×AB
AC/AP=AB/AC
∠A=∠A
∴△ACP∽△ABC
The angle is equal and the two sides are proportional
It is known that △ ABC is an equilateral triangle, P is a point in the triangle, PA = 3, Pb = 4, PC = 5
Rotate △ ABP clockwise about point B for 60 ° to get △ BCQ, connect PQ, ∵ - PBQ = 60 °, BP = BQ, ∵ - bpq is equilateral triangle, ∵ PQ = Pb = 4, and PC = 5, CQ = 4. In △ PQC, pq2 + QC2 = PC2, ∵ PQC is right triangle, ∵ - BQC = 60 ° + 90 ° = 150 °, and ∵ - APB = 150 °
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