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# Example of a Right Triangle

Question: In triangle ABC, angle A is 45^{o}, angle C is 90^{o} and AC = 3 times square root of 2. P is a moving point, moving along line CA but not overlap with points C and A. What is the range of BP?

- Solution:
- Because the angle C is 90
^{o}(given), - so, triangle ABC is a right triangle.
- In right triangle ABC, because the angle A = 45
^{o}(given), - so, angle B = 90
^{o}- angle A = 90^{o}- 45^{o}= 45^{o} - Because angle A = angle ABC,
- so, BC = AC = 3 × square root 2
- (In a triangle, congruent angles opposite congruent sides.)
- Since ABC is a right triangle,
- so, AB
^{2}= AC^{2}+ BC^{2} - and because AC = BC
- so, AB
^{2}= 2AC^{2}= 2 × (3 × square root of 2)^{2}= 2 × 9 × 2 = 36 = 6^{2} - so, AB = 6
- When point P move close to point C, then BP close to BC.
- When point P close to point A, then BP close to AB.
- Since point P is never overlap with point C or P (given),
- so, BC < BP < AB
- so, 3 × square root 2 < BP < 6