Is BCD a straight line?

Comments (3) on “Is BCD a straight line?”

1. ∠ECD  is 110°

   Step-by-step explanation:

   For a cyclic quadrilateral,

   ∠ABC + ∠AEC = 180°

   Therefore ∠AEC = 180° - 45° = 135°

   ∠CAE + ∠ACE + ∠AEC = 180° (Angles in a triangle)

   Therefore ∠CAE =  180 - 20 - 135 = 25°

   ∠CAE = ∠CDA = 25° (Base angles of isosceles triangle ΔCAD)

   ∠CED + ∠AEC = 180°(Sum of angles on a straight line)

   Therefore ∠CED + 135° = 180°

   ∠CED = 180° - 135° = 45°

   ∠ECD + ∠CED + ∠CDA = 180°

   Therefore ∠ECD  = 180° - (∠CED + ∠CDA)

   ∠ECD  = 180° - (45° + 25°) = 110°

   Therefore, ∠ECD  = 110°.

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2. see explanation

   Step-by-step explanation:

   (a)

   Since lines AB and CD are parallel then their slopes are equal

   y = - [tex]\frac{1}{2}[/tex] x + 14 is in slope- intercept form ( y = mx + c )

   with slope m = - [tex]\frac{1}{2}[/tex]

   Hence slope of CD = - [tex]\frac{1}{2}[/tex]

   D is the point (0, 7) ⇒ c = 7

   y = - [tex]\frac{1}{2}[/tex] x + 7 ← equation of CD

   (b)

   to find the x- intercept let y = 0, in the equation and solve for x

   - [tex]\frac{1}{2}[/tex] x + 7 = 0 ( multiply through by 2 )

   - x + 14 = 0 ⇒ - x = - 14 ⇒ x = 14 ← x- intercept of CD

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3. NO the line is slanted

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