1.8. Right triangles, Kiselev's Geometry, Book I. Planimetry, Givental Alexander.

 Exercises.

Prove Theorems:

103.* A line and a circle can have at most two points in common.

Proof: Assume that a line and a circle have three points (more than two points) in common. These points are distinct from each other. The line which joins these points cannot be straight because as we will pass from one point to another the line will change its direction. In other words, the line joining these points will be a broken line. By a line, the hypothesis implies a straight line. A contradiction has been reached.

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1.7. Inequalities in triangles, Kiselev's Geometry, Book I. Planimetry, Givental Alexander.

 A Memorable Theorem.

Theorem: In a triangle, each side is smaller than the sum of the other two sides.

Proof: We have the triangle ABC and we can see that AB is clearly smaller than the sum of sides BC and AC. Also, that side BC is smaller than the sum of AB and AC is clear. But that AC is smaller than the sum of AB and BC is not so clear.

We will prove the last statement. In triangle ABC extend AB to AD such that BD and BC are congruent. This implies that triangle BCD is isosceles. So ∠BCD = ∠BDC. Now ∠ACD ＞∠BCD, i.e. ∠ACD＞∠BDC. This implies that side AD is greater than AC, in other words, AD = AB + BD ＞AC. Or AB + BC＞AC. This proves the last statement. 

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1.6. Congruence tests for triangles, Kiselev's Geometry, Book I. Planimetry, Givental Alexander.

Exercises.

78. On each side of an equilateral triangle ABC, congruent segments AB', BC', and CA' are marked, and the points A', B', and C' are connected by lines. Prove that the triangle A'B'C' is also equilateral.

Proof: The triangle ABC is equilateral, so AB = AC and AB = BC. This implies that ∠ABC = ∠ ACB and ∠BAC = ∠BCA. So  ∠ABC = ∠BCA = ∠ACB. In other words ∠B'BC' = ∠A'AB' = ∠A'CC'. We already know that AB' = BC' = CA'. Further, AB = BC= CA implies that A'A = BB' = CC'. Thus, triangle A'CC' is congruent to C'BB' and also triangle A'CC' is congruent to A'AB', i.e. triangles A'CC', C'BB', and A'AB' are congruent to each other. We can conclude that A'C' = C'B' = A'B'. In other words, we have proved that triangle A'B'C' is equilateral.

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1.5. Isosceles triangles and symmetry, Kiselev's Geometry, Book I. Planimetry, Givental Alexander.

 Exercises.

63.* How many axes of symmetry can a quadrilateral have?

Solution: A quadrilateral can have at most four axes of symmetry.

66.* Two points A and B are given on the same side of a line MN. Find a point C on MN such that the line MN would make congruent angles with the sides of the broken line ACB.

Solution: 

The figure above gives the solution to the problem.

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1.4. Polygons and Triangles, Kiselev's Geometry, Book I. Planimetry, Givental Alexander.

 Some notes.

1. If a broken line is drawn such that the segment joining any two points of the broken line does not intersect the broken line, then the broken line is called closed.

- The figure above though is a counterexample.

2. For a broken line to be closed it is not necessary for it to be convex. Nor is it necessary for the broken line to be convex, that it be closed.

3. A segment joining two vertices of a polygon that have at least one point lying between them is called a diagonal of the polygon.

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 Exercises.

55. Prove that each diagonal either lies entirely in its interior, or entirely in its exterior.

Proof: Assume that a diagonal of a quadrilateral lies both in its interior and its exterior. This implies that the diagonal passes through a side of the quadrilateral The side connects two vertices. The diagonal will connect two opposite vertices. The resulting broken line ( which forms the boundary of the polygon ), though is not closed. A quadrilateral has to be closed. This implies that the polygon is not a quadrilateral. We have been so led to a contradiction.

56. Show that in any triangle, every two medians intersect.

Proof: Suppose that two medians of a triangle do not intersect. In other words, these two medians are parallel to each other. The medians join the midpoint of a side to its opposite vertex. There are two medians, so we should have two sides and two vertices. The resulting broken line which forms the boundary of the polygon though is not closed. A triangle has to be closed. The polygon so formed cannot be a triangle. A contraction has been so arrived at.

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1.2. Perpendicular Lines, Kiselev's Geometry, Book I. Planimetry, Givental Alexander.

 Exercises.

Prove:

40. If at a point O of the line AB (in the above figure) two congruent angles AOD and BOC are built on the opposite sides of AB, then their sides OD and OC form a straight line.

Proof: Assume that OD and OC do not form a straight line.

Then ∠AOC, ∠AOD, and ∠AOC, ∠BOC are not supplementary pairs, i.e. the sums ∠AOC + ∠AOD and ∠AOC + ∠BOC cannot be equal. In other words ∠AOD and ∠BOC cannot be equal. We have reached a contradiction.

41. If from the point O (in the same above figure) rays OA, OB, OC, OD are constructed in such a way that ∠AOC = ∠DOB and ∠AOD = ∠COB, then OB is the continuation of OA, and OD is the continuation of OC.

Proof: Now ∠AOC = ∠DOB and ∠AOD = ∠COB.

We know that ∠AOC + ∠DOB +∠AOD + ∠COB = 360°, because the sum forms a full angle. This implies that 2∠AOC + 2∠AOD = 360°. In other words, ∠AOC + ∠AOD = 180°, i.e. ∠AOC and ∠AOD are supplementary. Thus, OD is the continuation of  OC.

In a similar way we can prove that OB is the continuation of OA. Both cases are now proved.

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1. The Straight Line, 1. Angles, Kiselev's Geometry, Book I. Planimetry, Givental Alexander.

 Exercises.

27*. Using only compass, construct a 1° arc on a circle, if a 19° arc of this circle is given.

Solution: 

In the above figure we have a circle with center A. We have the arc BB' of 19 circular degrees. The angle 𝛂 having 19 angular degrees. We pass two lines one passing through the points AB and another passing through the points AB'. Taking the radius as AC we draw a circular arc CD which intersects the line AB' in point D. Now join the segment CD. The line CD is now a projection of the line BB', where point A is the source of light and ∠ B'AB has its sides AB' and AB as the rays forming the very ends of the light that is being emitted out of the source A. We can divide the line CD into 19 equal parts in this way: Pass a line through point D making an acute angle 𝜸 in the anti-clockwise direction with CD as its leg. Then pass a line through point C making an angle equal to 𝜸 in the same direction with DC as its leg. 

Start with point C. Put the pin of the compass onto C and the pencil point in the direction of the line passing through C and making the angle 𝜸. Keep the distance between the pin and the pencil constant and mark points in the direction of the line passing through C, with each point being marked becoming the point where the pin of the compass will be placed for the next point to be marked. Mark 19 points in such way. 

Now put the pin on the point D and in the direction of the line passing through D and making the angle equal to 𝜸, keeping the distance same mark 19 points as were marked above. 

Now we have these two lines parallel to each other and divided into 19 equal parts. Start from C. Join point C with the last marked point on the line passing through D. Then join the next marked point next to C with the marked second last point next to the last point. Establish such a correspondence. And join the points. The lines which join one point to its corresponding point will divide the segment CD into 19 equal parts. Join each of the points of division with the center point A. 

The arc CD has now been divided into 19 equal parts. Thus, its corresponding arc BB' has also been divided into 19 equal parts. Select any one of those 19 equal parts. The arc that congruent to this one part will have measure 1°. We have successfully constructed an arc of 1° on the circle to which the corresponding angle is 𝜷 with A' as its vertex.

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