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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
1 year, 11 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.

Gwen Roberts - 1 year, 11 months ago

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Please look at my solution. Thank you.

Gwen Roberts - 1 year, 11 months ago

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