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# Geo help

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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

Feb 19, 2018

### 2+0 Answers

#1
+99592
+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

Feb 20, 2018
#2
+21990
+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 $$\angle C+\angle D$$?

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

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

Feb 20, 2018
edited by heureka  Feb 20, 2018