in a quadrilateral ABCD, BC is parallel to AD. AB = BC, 13CD = 20BC and 34AB = 13AD. H is a point on AD such that CH is perpendicular to AD. Prove that 5CH = 3CD

Answer 1

Proof:

Since BC is parallel to AD, we know that the angles between AB and BC and between AD and CH are equal.

Let angle ABC = x

Then angle ADH = x

Since AB = BC, we know that angle ABC = angle BCD = x

Since 13CD = 20BC, we know that angle BCD = (20/13)x

Since 34AB = 13AD, we know that angle ABC = (13/34)x

Since CH is perpendicular to AD, we know that angle ADH = 90°

Therefore, (20/13)x + (13/34)x + 90° = 180°

Simplifying, we get 5x + 3x + 90° = 180°

Solving for x, we get x = 30°

Therefore, angle ABC = angle BCD = 30°

Since CH is perpendicular to AD, we know that angle CHD = 90°

Therefore, angle CHD + angle BCD = 90° + 30° = 120°

Since angle CHD = angle BCD, we know that 5CH = 3CD