Point O is any point in the triangle ABC, connecting Ao, Bo and Co. proof: ab + AC > ob + OC AB + BC + AC > OA + ob + OC

In △ abd, AB + ad > BD, BD = ob + OD, ∧ AB + ad > ob + od in △ ODC, OD + CD > OC, ∧ AB + AD + od + CD > ob + DD + OC, ∧ AD + CD = AC ∧ AB + AC > OC + ob, there are ab + BC > OA + OC, AC + BC > OA + ob, the three formulas are: 2 (AB + BC + AC) > 2 (OA + ob + OC) ∧ AB + BC + AC ∧

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1. As shown in the figure, O is any point in the triangle ABC, connecting Ao, Bo and Co
2. As shown in the figure, point O is a point outside △ ABC. Take a ', B', C 'on the ray OA, ob, OC respectively, so that oa'oa = ob'ob = oc'oc = 3. Connect a'B', b'c ', c'a', and whether the obtained △ a'b'c 'is similar to △ ABC? Prove your conclusion
3. Given that O is the outer center of △ ABC, e is the inner point of triangle, and OE = OA + ob + OC, it is proved that AE is perpendicular to BC
4. In the triangle ABC, if AA CC + BC = BB, then a =?
5. Three sides ABC of triangle, find aa-bb-cc-2ab < 0
6. The triangle ABC must satisfy π / 3
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8. AAA + BBB + CCC = CBBC question; a =? B =? C =?
9. AAA+BBB+CCC=CBBC What does a stand for? What does B stand for
10. AAA+BBB-CCC=?DDD A
11. Take any point O in the triangle ABC, connect Ao, Bo, CO respectively, and extend the opposite edge to a ', B', C '. Prove: OA' / AA '+ ob' / BB '+ OC' / CC '= 1
12. As shown in the figure, in the triangle ABC, the angle BAC is equal to 90 degrees, ad is perpendicular to BC, and BD square = BD times BC
13. As shown in the figure, in the triangle ABC, CD is the height on the side of AB, and the square of CD = ad * BD. try to explain that the triangle ABC is a right triangle
14. In the triangle ABC, if the angle a = 120 degrees, ab = 4, AC = 2, then SINB =?
15. In the triangle ABC, the angle a = 120 °, ab = 4, AC = 2, then the value of SINB is () Originally there was no picture
16. In △ ABC, ab = AC = 10, BC = 16, find the value of tanb
17. In △ ABC, ab = AC = 10, BC = 16, find the value of tanb
18. As shown in the figure, in △ ABC, ab = AC, D, e are two points on the straight line BC, and ab & # 178; = DB × CE, if ∠ BAC = 40 °, calculate the degree of ∠ DAE
19. As shown in the figure, in △ ABC, D and E are on the straight line BC. (1) if AB = BC = AC = CE = BD, calculate the degree of ∠ EAC; (2) if AB = AC = CE = BD, ∠ DAE = 100 °, calculate the degree of ∠ EAC
20. Given that D and E are two points on the edge BC of △ ABC, and ∠ bad = ∠ C, ∠ DAE = ∠ EAC, it is proved that BD ratio AB is equal to de ratio CE