Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 sides...is it true...?? i am not able to prove it.

share|improve this question
1  
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
1  
Hint: Compare $\triangle B^\prime A C^\prime$ to $\triangle B^\prime B A^\prime$. –  Blue Nov 29 '13 at 7:17

3 Answers 3

up vote 1 down vote accepted

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

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

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

Therefore,

  • $\triangle B^\prime AC^\prime$ 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 $150^\circ$.

Therefore,

  • $|\angle A^\prime BB^\prime| = 150^\circ - 60^\circ = 90^\circ$, 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. $\square$

share|improve this answer

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.

share|improve this answer

$\bigtriangleup$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
share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.