Point O is a point outside the triangle ABC. Take a'b'c 'on the ray OA, ob, OC respectively, so that OA' / OA = ob '/ ob = OC' / OC = 3, connecting a'B ', b'c', c'a '. How many times of the area of triangle a'b'c' is that of triangle ABC

Point O is a point outside the triangle ABC. Take a'b'c 'on the ray OA, ob, OC respectively, so that OA' / OA = ob '/ ob = OC' / OC = 3, connecting a'B ', b'c', c'a '. How many times of the area of triangle a'b'c' is that of triangle ABC

The area of triangle a'b'c 'is nine times that of triangle ABC
(2 points)
It is proved that from the known OA ′ / OA = OC ′ / OC = 3, ∠ AOC = ∠ a ′ OC ′
(4 points)
A 'C' / AC = OA '/ OA = 3, B' C '/ BC = 3, a' B '/ AB = 3
S△A'B'C'/S△ABC=9/1

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

It is proved by △ a ′ B ′ C ′∽ ABC. (2 points) that: from the known OA ′ OA = OC ′ OC = 3, ∠ AOC = ∠ a ′ OC ′∽ AOC ∽ a ′ OC ′, (4 points) ‖ a ′ C ′ AC = OA ′ OA = 3, similarly, B ′ C ′ BC = 3, a ′ B ′ AB = 3. (6 points) ‖ a ′ C ′ AC = B ′ C ′ BC = a ′ B ′ ab. (7 points) ‖ △ a ′ B ′ C ′∽ ABC. (8 points)

As shown in the figure, take any point O in the triangle ABC to connect OA, ob, OC, a'b'c ', which are OA, OB.OC It is proved that △ ABC is similar to △ a'b'c '

Draw vertical lines from O, a ', B' and C 'to AB, AC and BC respectively
It can be proved that a'B '= 1 / 2 * AB, a'c' = 1 / 2 * AC, b'c '= 1 / 2 * BC
Namely: a'B ': a'c': b'c '= AB: AC: BC
It is concluded that △ ABC is similar to △ a'b'c

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

It is proved by △ a ′ B ′ C ′∽ ABC. (2 points) that: from the known OA ′ OA = OC ′ OC = 3, ∠ AOC = ∠ a ′ OC ′∽ AOC ∽ a ′ OC ′, (4 points) ‖ a ′ C ′ AC = OA ′ OA = 3, similarly, B ′ C ′ BC = 3, a ′ B ′ AB = 3. (6 points) ‖ a ′ C ′ AC = B ′ C ′ BC = a ′ B ′ ab. (7 points) ‖ △ a ′ B ′ C ′∽ ABC. (8 points)

The proof pi / 3 of ABC in triangle ABC

aA+bB+cC=aπ-aB-aC+bπ-bA-bC+cπ-cA-cB
=π(a+b+c)-[A(b+c)+B(a+c)+C(a+b)
aA+bB+cC+[A(b+c)+B(a+c)+C(a+b)=π(a+b+c)
b+c>a a+c>b a+b>c A(b+c)+B(a+c)+C(a+b)>aA+bB+cC
π(a+b+c)>2(aA+bB+cC)
(aA+bB+cC)/(a+b+c)c c-a>b
aA+bB+cC>πa+cB+bC>πa
Similarly, AA + BB + CC > π B
Similarly, AA + BB + CC > π C
3(aA+bB+cC)>π(a+b+c)
(aA+bB+cC)/(a+b+c)>π/3