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AB is the diameter of ⊙ o, C is on ⊙ o, BP is the middle line of △ ABC, BC = 3, AC = 62, find the length of BP
∵ AB is the diameter of ⊙ o, ∵ C = 90 ° and ∵ BP is the middle line of △ ABC, ∵ CP = 12ac = 32. In the right angle △ BCP, BP = PC2 + BC2 = (32) 2 + 32 = 33
As shown in Figure 3, it is known that in RT △ ABC, ∠ C = 90 °, ab = 2Ac = 2 times root sign 3, ad bisects ∠ BAC, CD = 1, and finds the length of AD
2AC=2√3
So AC = √ 3
CD=1
So ad = √ (3 + 1) = √ 4 = 2
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Given the right angle △ ABC, angle c = 90 °, ab = 2Ac, BC = 2cm, then AC =?
Let AC = x, because AB = 2Ac, so AB = 2x, and because it is a right triangle, so AC = 2 can be calculated according to Pythagorean theorem
In RT △ ABC, ad is the middle line on the hypotenuse BC, ad = 3cm, try to find; (1) the length of BC; (2) if AB = 4cm, find the area of △ ABC
In ∵ RT △ ABC, ad is the median line on the hypotenuse BC
∴AD=BD=DC=3
∴BC=6
And ∵ AB = 4 (known) according to Pythagorean theorem
∴AC=2√5
Triangle area s = 1 / 2Ab = 1 / 2 × 4 × 2 √ 5 = 4 √ 5
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