back to geometry video lessons

Example of a Right Triangle

Question: In triangle ABC, angle A is 45o, angle C is 90o 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 90o (given),
so, triangle ABC is a right triangle.
In right triangle ABC, because the angle A = 45o (given),
so, angle B = 90o - angle A = 90o - 45o = 45o
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, AB2 = AC2 + BC2
and because AC = BC
so, AB2= 2AC2 = 2 × (3 × square root of 2)2 = 2 × 9 × 2 = 36 = 62
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