# How should I approach this problem?

Let $$ABC$$ be a triangle with $$AB = AC$$ and $$\angle BAC = 30^{\circ}$$, Let $$A'$$ be the reflection of $$A$$ in the line $$BC$$; $$B'$$ be the reflection of $$B$$ in the line $$CA$$; $$C'$$ be the reflection of $$C$$ in line $$AB$$, Show that $$A'B'C'$$ is an equilateral triangle.

Any help will be appreciated.

Note by Shashank Rammoorthy
2 years, 4 months ago

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Draw the figure and notice that connecting points BCB'C' results in a rectangle because the diagonals (CC' and BB' bisect each other). AA' contains a segment that bisects B'C' and also bisects rectangle BCB'C' vertically in to 2 smaller congruent rectangles, whose diagonals are congruent, A'C' = A'B'. Since A'B is half of A'C', all three sides of triangle A'B'C' are equal and it is equilateral.

- 2 years, 4 months ago

Please look at my solution. Thank you.

- 2 years, 4 months ago