# Couple of elementary geometry problems

1. In triangle ABC AB=BC=4, angle ABC=30 degrees. On the outside of sides AB and BC, equilateral triangles ABD and BCF are constructed. Lines AF and CD intersect in point O. Find angles of the triangle AOC and distance between lines AC and DF.

2. Bisectors of angles A and B of parallelogram ABCD intersect in point K on the side CD. Perimeter of ABCD is 45, and difference of perimeters of triangles BCK and ADK is 3. Find sides of the parallelogram and lengths of segments AK and BK.

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#### skipjack

Forum Staff
Have you drawn accurate diagrams? If so, what progress have you made from there?

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Here are the diagrams: The first diagram is pretty accurate, the second one is not to scale.

As far as progress:

1. Some observations:
DF is parallel to AC
Lines defined by DA and FC intersect at the right angle
Triangles DFO and CAO are similar
Many of the angles can be easily calculated, but I can't seem to get to the ones that are truly needed, such as ADC

2. I'm pretty stuck on this one - wrote the relationships between the perimeters algebraically, but too many unknowns

I got the the second one actually. Leaving out the formal proofs:
1. Bisectors of the parallelogram intersect at right angle
2. Denote the intersection point of line AK and BC as L; Intersection of BK and AD as M
3. ABLM is a rhombus since its diagonals cross at the right angle
4. CD divides BL and AM in half, meaning that AB=CD=15; AD=BC=7.5
5. ABK is a right triangle. Denoting BK as x: 15^2=x^2+(x+3)^2 => x=9

Ok, figured out the first part of problem 1, but still struggling with the distance part.

1. Angle BAC equals angle BCA = (180-30)/2=75 degrees
2. Angle DBC=30+60=90 degrees (30 is given, and 60 because ADB is equilateral, as given)
3. DBC is isosceles, as given, hence angle DCB=45 degrees
4. Angle ACO=CAO=75-45=30 degrees, hence angle AOC=120 degrees

I tried to post the diagrams as dropbox links, but they are not showing for some reason.

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#### skipjack

Forum Staff 1. Your work is correct. Note that triangles BCD, BFA are congruent, which can alternatively be used. To find the distance between AC and DF, draw a perpendicular from F to AC extended. It is one leg of an isosceles right-angled triangle that has FC as its hypotenuse.

2. ABLM is a rhombus, but your stated reason is insufficient. You can alternatively proceed (without needing any construction) by proving that the two small triangles are isosceles.

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