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 \(\angle C+\angle D\)?
+128408 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
