As shown in the figure: O is any point in △ ABC, a '. B'. C 'is OA.OB.OC Is triangle ABC similar to triangle a'b'c '? Why?

As shown in the figure: O is any point in △ ABC, a '. B'. C 'is OA.OB.OC Is triangle ABC similar to triangle a'b'c '? Why?

It has nothing to do with O
Similarity is inevitable, the median line is parallel to the bottom, then we can prove the similarity by using parallel or AAA~

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

As shown in the figure ∵ OA '/ OA = ob' / ob = 1 / 2, ∵ a'ob '= ∵ AOB ∵ a'ob' ∵ AOB, ∵ a'b'o = ∵ ABO, we can get ∵ c'b'o = ∵ CBO, ∵ a'b'o + ∵ c'b'o = ∵ ABO + ∵ CBO, that is ∵ a'b'c '= ∵ ABC and ∵ a'B' = 1 / 2Ab, c'b '= 1 / 2CB (triangle median line theorem) ∵ a'B' / AB = c'b '/ CB

Let OA = a ob = TB OC = 1 / 3 (a + b) prove that ABC is collinear when t is what value
It's all vectors

See figure
When at least one of a and B is a zero vector, t takes any value
When a and B are not zero vectors and a and B are collinear, t takes any value
When a and B are not zero vectors and a and B are not collinear, t is 1 / 2

Let m and n be the middle points of BC and EF respectively. It is proved that Mn is perpendicular to E

Proof: because CE is perpendicular to AB, BF is perpendicular to AC,
So triangle BCE and triangle BCF are right triangles, and BC is the common hypotenuse,
And because m is the midpoint of BC,
So me = MF = BC / 2,
Because me = MF, n is the midpoint of EF,
So Mn is perpendicular to ef