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In triangle \(ABC,\) let \(I\) be the intersection of angle bisectors \(\overline{AD}\) and \(\overline{BE}.\) Prove that    \(\angle DIB = 90^\circ - \frac{\angle BCA}{2}.\)

 

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Thank you!

 Feb 12, 2020

Best Answer 

 #1
avatar+24430 
+2

In triangle \(ABC\),  let \(I\) be the intersection of angle bisectors \(\overline{AD}\) and  \(\overline{BE}\).
Prove that \( \angle DIB = 90^\circ - \dfrac{\angle BCA}{2}\).  


\(\begin{array}{rcll} \begin{array}{|rcll|} \hline \mathbf{\text{In AIB:}} \\ \hline 180^\circ-x + \alpha + \beta &=& 180^\circ \\ -x + \alpha + \beta &=& 0 \\ \mathbf{x} &=& \mathbf{\alpha + \beta} \qquad (1) \\ \hline \end{array} & \begin{array}{|rcll|} \hline \mathbf{\text{In BID:}} \\ \hline x + D + \beta &=& 180^\circ \\ \mathbf{180^\circ-D} &=& \mathbf{x+\beta} \qquad (2) \\ \hline \end{array} \\ & \begin{array}{|rcll|} \hline \mathbf{\text{In EIA:}} \\ \hline x + E + \alpha &=& 180^\circ \\ \mathbf{180^\circ-E} &=& \mathbf{x+\alpha} \qquad (3) \\ \hline \end{array} \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\text{In EIDC:}} \\ \hline (180^\circ-D)+(180^\circ-E)+(180^\circ-x) + C &=& 360^\circ\\ \boxed{ \mathbf{180^\circ-D=x+\beta}\\ \mathbf{180^\circ-E=x+\alpha}} \\ x+\beta+x+\alpha+(180^\circ-x) + C &=& 360^\circ \\ x+\beta+x+\alpha+180^\circ-x + C &=& 360^\circ \quad | \quad - 180^\circ \\ x+\beta+x+\alpha -x + C &=& 360^\circ - 180^\circ \\ x+\beta+x+\alpha -x + C &=& 180^\circ \\ x+\beta +\alpha + C &=& 180^\circ \\ x+\alpha+\beta + C &=& 180^\circ \quad | \quad \mathbf{\alpha+\beta=x} \\ x+x + C &=& 180^\circ \\ 2x + C &=& 180^\circ \\ 2x &=& 180^\circ -C \quad | \quad : 2 \\ \mathbf{x} &=& \mathbf{90^\circ -\dfrac{C}{2}} \\ \hline \end{array}\\\)

 

laugh

 Feb 12, 2020
 #1
avatar+24430 
+2
Best Answer

In triangle \(ABC\),  let \(I\) be the intersection of angle bisectors \(\overline{AD}\) and  \(\overline{BE}\).
Prove that \( \angle DIB = 90^\circ - \dfrac{\angle BCA}{2}\).  


\(\begin{array}{rcll} \begin{array}{|rcll|} \hline \mathbf{\text{In AIB:}} \\ \hline 180^\circ-x + \alpha + \beta &=& 180^\circ \\ -x + \alpha + \beta &=& 0 \\ \mathbf{x} &=& \mathbf{\alpha + \beta} \qquad (1) \\ \hline \end{array} & \begin{array}{|rcll|} \hline \mathbf{\text{In BID:}} \\ \hline x + D + \beta &=& 180^\circ \\ \mathbf{180^\circ-D} &=& \mathbf{x+\beta} \qquad (2) \\ \hline \end{array} \\ & \begin{array}{|rcll|} \hline \mathbf{\text{In EIA:}} \\ \hline x + E + \alpha &=& 180^\circ \\ \mathbf{180^\circ-E} &=& \mathbf{x+\alpha} \qquad (3) \\ \hline \end{array} \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{\text{In EIDC:}} \\ \hline (180^\circ-D)+(180^\circ-E)+(180^\circ-x) + C &=& 360^\circ\\ \boxed{ \mathbf{180^\circ-D=x+\beta}\\ \mathbf{180^\circ-E=x+\alpha}} \\ x+\beta+x+\alpha+(180^\circ-x) + C &=& 360^\circ \\ x+\beta+x+\alpha+180^\circ-x + C &=& 360^\circ \quad | \quad - 180^\circ \\ x+\beta+x+\alpha -x + C &=& 360^\circ - 180^\circ \\ x+\beta+x+\alpha -x + C &=& 180^\circ \\ x+\beta +\alpha + C &=& 180^\circ \\ x+\alpha+\beta + C &=& 180^\circ \quad | \quad \mathbf{\alpha+\beta=x} \\ x+x + C &=& 180^\circ \\ 2x + C &=& 180^\circ \\ 2x &=& 180^\circ -C \quad | \quad : 2 \\ \mathbf{x} &=& \mathbf{90^\circ -\dfrac{C}{2}} \\ \hline \end{array}\\\)

 

laugh

heureka Feb 12, 2020
 #2
avatar+109740 
+1

Nice, heureka  !!!

 

 

cool cool cool

CPhill  Feb 12, 2020
 #3
avatar+24430 
+1

Thank you, CPhill !

 

laugh

heureka  Feb 12, 2020

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