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Triangles ABC and ABD are isosceles with \(AB=AC=BD\), and \(\overline{BD}\) intersects \(\overline{AC}\) at \(E\). If \(\overline{BD}\perp\overline{AC}\), then what is the value of \(\angle C+\angle D\)?

 

https://latex.artofproblemsolving.com/8/3/6/836d27bee8c65b6b978cfc8c1ef0662feec4c994.png

mathtoo  Feb 19, 2018
 #1
avatar+88891 
+1

Angle C  =  ACB      Angle D  = ADB

 

BAD  = ADB         ABC  = ACB

 

 

DBA  + BAC   =  90    ⇒   DBA  = 90 - BAC 

ACB + DBC  = 90    ⇒      DBC =  90 - ACB 

DAC + ADB  = 90   ⇒       DAC  = 90 - ADB       

 

DAC + BAC  = BAD                      DBA + DBC = ABC

DAC + BAC  = ADB                      DBA + DBC =  ACB

[90 - ADB] + BAC  = ADB             DBA + [90 - ACB]  = ACB

90 + BAC = 2ADB   (1)                  90 + DBA  = 2ACB

                                                      90 + [90 - BAC]  = 2ACB

                                                     180  - BAC  = 2ACB

                                                      BAC  = 180 - 2ACB   (2)

 

Sub  (2)  into (1)

90 + [180 - 2ACB]  = 2ADB

270 =  2ACB + 2ADB

135  = ACB + ADB

135  =  ∠C +  ∠ D

 

 

cool cool cool

CPhill  Feb 20, 2018
 #2
avatar+20009 
+1

Triangles ABC and ABD are isosceles with

\(AB=AC=BD\),
and\( \overline{BD}\) intersects\( \overline{AC}\) at \(E\).
If \(\overline{BD}\perp\overline{AC}\), then what is the value of \(\angle C+\angle D\)?

 


\(\angle D = BDA = DAB \\ \angle C = BCA = CBA \\ \angle \alpha = DBA\\ \angle \beta = BAC \\ \angle \alpha +\angle \beta = 90^{\circ}\)

 

\(\begin{array}{|rcll|} \hline \angle \alpha &=& 180^{\circ} - 2\times \angle D \\ \angle \beta&=& 180^{\circ} - 2\times \angle C \\ \angle \alpha +\angle \beta &=& 180^{\circ} - 2\times \angle D+180^{\circ} - 2\times \angle C \quad & | \quad \angle \alpha +\angle \beta = 90^{\circ}\\ 90^{\circ} &=& 180^{\circ} - 2\times \angle D+180^{\circ} - 2\times \angle C \\ 90^{\circ} &=& 360^{\circ} - 2\times (\angle C + \angle D) \\ 2\times(\angle C + \angle D) &=& 360^{\circ} - 90^{\circ} \\ 2\times(\angle C + \angle D) &=& 270^{\circ} \\ \mathbf{ \angle C + \angle D } & \mathbf{=}& \mathbf{135^{\circ}} \\ \hline \end{array}\)

 

laugh

heureka  Feb 20, 2018
edited by heureka  Feb 20, 2018

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