Six points are chosen on the sides of an equilateral triangle ABC: A1, A2 on BC, B1, B2 on CA and C1, C2 on AB, such that they are the vertices of a convex hexagon A1, A2, B1, B2, C1, C2 with equal side lengths. Prove that the lines A1 B2, B1 C2 and C1 A2 are concurrent.
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In ΔAB2C1 and ΔBC2A1
∠A=∠B
If, ∠BC2A1 > ∠AB2C1
Then, ∠BA1C2 < ∠AC1B2
But |A1C2|=|B2C1|
Therefore, the law of sines implies,
|BA1|>|AC1|⇒|BC2|<|AB2|
On the other hand we have,|BC2|+|AC1|=|AB2|+|CB1|. Therefore,
|BC2|<|AC1|⇒|AC1|>|CB1|
By the similar arguement, we have:-
|BA1|>|AC1|⇒|AC1|>|CB1|
⇒|CB1|>|BA1|
Since, the contradiction shows that |BA1|=|AC1|=|CB1|
Thus, the three triangles ΔAB2C1, ΔBC2A1 and ΔCA2B1 are congruent.
This implies that the triangle, ΔA2B2C2 is equilateral and A1B2, B1C2 and C1A2 are it's heights. Therefore, they are concurrent.
Hope this helps you!! :)
∠A=∠B
If, ∠BC2A1 > ∠AB2C1
Then, ∠BA1C2 < ∠AC1B2
But |A1C2|=|B2C1|
Therefore, the law of sines implies,
|BA1|>|AC1|⇒|BC2|<|AB2|
On the other hand we have,|BC2|+|AC1|=|AB2|+|CB1|. Therefore,
|BC2|<|AC1|⇒|AC1|>|CB1|
By the similar arguement, we have:-
|BA1|>|AC1|⇒|AC1|>|CB1|
⇒|CB1|>|BA1|
Since, the contradiction shows that |BA1|=|AC1|=|CB1|
Thus, the three triangles ΔAB2C1, ΔBC2A1 and ΔCA2B1 are congruent.
This implies that the triangle, ΔA2B2C2 is equilateral and A1B2, B1C2 and C1A2 are it's heights. Therefore, they are concurrent.
Hope this helps you!! :)
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