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Let $ABC$ be a triangle with $AB = AC $ and $angle BAC = 30.$ 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 the line AB. Show that $(A',B',C')$ form the vertices of an equilateral triangle.

I have been able to solve this..but I think the new triangle sides would be parallel to the older it true...?? i am not able to prove it.

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I think the problem statement needs a correction, it must be $AB=AC$. – hhsaffar Nov 29 '13 at 6:34
@hhsaffar ya u r correct.. – maths lover Nov 29 '13 at 6:45
It's true that $\overline{BC} \parallel \overline{B^\prime C^\prime}$, but the other new edges are not parallel to their counterparts. They can't be: the other two original edges bound an angle of $30^\circ$, while the other two new edges bound an angle of $60^\circ$. – Blue Nov 29 '13 at 7:16
Hint: Compare $\triangle B^\prime A C^\prime$ to $\triangle B^\prime B A^\prime$. – Blue Nov 29 '13 at 7:17
up vote 1 down vote accepted

Define $d := |\overline{AB}| = |\overline{AC}|$.

By the reflection of $B$ to $B^\prime$ and $C$ to $C^\prime$,

  • $|\angle B^\prime A C| = |\angle CAB| = |\angle C^\prime A B| = 30^\circ$
  • $|\overline{AB^\prime}| = |\overline{AC}| = |\overline{AB}| = |\overline{AC^\prime}| = d$


  • $\triangle B^\prime AC^\prime$ is an isosceles right triangle with legs $d$.
  • $\triangle BAB^\prime$ is isosceles with vertex angle $60^\circ$, hence it is equilateral, with all sides of length $d$ and all angles of measure $60^\circ$.

By the reflection of $A$ to $A^\prime$,

  • $\square ABA^\prime C$ is a rhombus with side lengths $d$ and acute angle measure $30^\circ$; its obtuse angle has measure $150^\circ$.


  • $|\angle A^\prime BB^\prime| = 150^\circ - 60^\circ = 90^\circ$, so that $\triangle A^\prime BB^\prime$ is an isosceles right triangle with legs $d$; likewise for $\triangle A^\prime CC^\prime$.

Thus, $\triangle B^\prime A C^\prime \cong \triangle C^\prime CA^\prime \cong \triangle A^\prime BB^\prime$, so that $\overline{B^\prime C^\prime} \cong \overline{C^\prime A^\prime} \cong \overline{A^\prime B^\prime}$, as desired. $\square$

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Only $BC$ is parallel to $B'C'$. To see that $A'C'$ is not parallel to $AC$, it suffice to prove that $\angle CAC'+\angle A'CA\not=180$. With some angle chasing, you will find that $\angle CAC'=60$ and $\angle A'CA=105$.

The argument for $A'B'$ not parallel to $AB$ is similar.

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$\bigtriangleup$A'B'C' is not equilateral necessarily but it is isoscalets surely

  • now B'C=BC=C'B and BA'=A'C[as AB=AC A'D is the perpendicular bisector of BC]
  • and $\angle$C'BA'=$\angle$B'CA' so $\bigtriangleup$C'BA'$\cong$$\bigtriangleup$B'CA'
  • so A'C'=A'B'
  • BUT if you now consider $\bigtriangleup$A'B'C' to be equilateral then A'B'=C'B'AND we see B'C=AB' and C'A=AC=CA' then $\Delta$B'AC'$\cong$$\Delta$B'CA' then we will have $\angle$B'CA'=$\angle$B'AC' but $\angle$B'AC'=90 and $\angle$B'CA'=135
  • so we have a contradiction now!!enter image description here
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