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

 Jul 2, 2019
edited by Guest  Jul 2, 2019
 #1
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Let angle BAE  = 1  Let angle DAE  = 2   Let angle ABE = 3    Let angle ADE = 4  Let angle EBC = 5

Let angle ECB  = 6

So.....since triangle ABC  and triangle  ABD are isosceles....we have that

3 + 5  = 6       (1)

1 + 2  = 4       (2)

 

3  = 90° - 1     (3)

2 = 90° - 4     (4)

5 = 90° - 6      (5)

 

 

Sub (3), (4) and , (5)     into   (1) and (2)

 

(90° - 1) + (90° - 6)  = 6      ⇒  180° - 1  = 2*6      (7)

1  +  (90° -4)  = 4  ⇒            90° + 1  = 2*4         (8)

 

Add (7) and (8)  and we have that

 

270°  = 2*4 + 2*6

270° = 2 (4 + 6)  divide both sides by  2

135°  = 4 + 6  =  Angle ADE + angle ECB =  angle C + angle D

 

 

cool cool cool

 Jul 2, 2019

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