From points C and D, lying on one of the sides of a given acute angle, perpendiculars to this side are drawn

From points C and D, lying on one of the sides of a given acute angle, perpendiculars to this side are drawn, intersecting the second side of the angle at points A and B, respectively. a) Prove that AC || BD b) Find the angle CAB if the angle ABD = 55 degrees

1. Let us prove that AC || BD.

For straight AC and BD straight CD is the secant. ∠ACD = 90 ° and ∠BDE = 90 °. These angles are appropriate.

And since ∠ACD = ∠BDE, AC || BD.

2. Find ∠CAB.

∠CAB and ∠ABD are internal one-way for parallel lines AC and BD and secant AB.

The sum of one-sided angles is equal to 180 ° based on the parallelism of straight lines. Means,

∠CAB + ∠ABD = 180 °;

∠CAB = 180 ° – ∠ABD;

∠CAB = 180 ° – 55 °;

∠CAB = 125 °.

Answer: ∠CAB = 125 °.