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.

#Geometry #RMO #India #MathProblem #1998

Easy Math Editor

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Comments

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 - 5 years, 6 months ago

Please look at my solution. Thank you.

Gwen Roberts - 5 years, 6 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$