Solving Segment Proofs

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In geometry, a line segment is a division of a line that is bounded by two end points, and contains every point on the line between its end points. Examples of a segments include the circles. More generally, when the two end points are both vertices of a polygon, the line segment is either an edge (of that polygon) if they are adjacent vertices, or otherwise a diagonal.Let we see about solving segment proofs. Source –Wikipedia.
Angles in a Circle with Solving Proofs

Angles in the same segment of a circle are equal. Find the solving proofs.

Segments of circle:

The region in the interior of a circle is divided into two parts by the chord AB (see the diagram )These two regions are called segments. The segment which contains that the centre is the major segment and the smaller region is the minor segment. A diameter divides the circumference into two equal parts. Each of these is called a semicircle.

Let us now determine with the case when an arc AB is a semicircle (see the diagram A). As done in theorem 6, we join CO and produce it to P. We note that.

m angle of AOP + m angle ...
... of BOP = 180°

m angle of AOB = 180°

But m angle of AOB = 2m angle of ACB

m angle of ACB = ½ m angle of AOB

= `1/2` × 180°= `180/2` = 90° = a right angle.

Prove that angle in a minor segment is obtuse.Find the solving proofs.

Solution:

Let angle of ACB be an angle formed in a minor segment of a circle with centre O. We have to prove that angle of ACB is obtuse.
Now m angle of ACB = ½ reflex m angle of AOB but reflex m angle of AOB 180°

½ reflex m angle of AOB ½ × 180o or ½ reflex m angle of AOB 90°

Hence m angle of ACB 90°

That is m angle of ACB is an obtuse angle in the form of minor segment.
Angles in Solving the Alternate Segment Proofs

The angle between the tangent and a chord passing through the point of contact is equal to the angle in the alternate segment.

Given: TAT′ is a tangent and AB is a chord through the point of contact A.

To prove: m angle of TAB = m angle of ADB, m angle of T′AB = m angle of ACB .

Construction: Draw diameter AE and join BE

Proof : m angle of T′AE = 90°

m angle of T′AE = m angle of T′AB + m angle of BAE (Parts of angle)

∴ m angle of T′AB +m angle of BAE = 90°...................................................................... (1)

m angle of EBA = 90° (Angle in the semicircle)

m angle of AEB+m angle of BAE = 90° (ABE right angled Triangle) .....................(2)

From (1) and (2)

m angle of T′AB+m angle of BAE = m angle of AEB+m angle of BAE

m angle of T′AB = m angle of AEB (Removing common angle of BAE) ................(3)

m angle of AEB =m angle of ACB (Angles in the same segment) ...........................(4)

From (3) and (4)

m angle of T′AB = m angle of ACB

m angle of TAB + m angle of T′AB = 180° (TAT′ is a line)......................................... (5)

m angle of ACB + m angle of ADB = 180° (ABCD is cyclic).................................. (6)

From (5) an d(6)

m angle of TAB + m angle of T′AB = m angle of ACB +m angle of ADB

But m angle of T′AB =m angle of ACB

m angle of TAB = m angle of ADB

Hence the theorem is proved.

Segment Addition Postulate is an important subject that is educated to all academic levels. Many students struggle to write their report on Segment Addition Postulate, report on Segment Addition Postulate term paper or Segment Addition Postulate analysis.

The Segment Addition Postulate is repeatedly used in geometric proofs to designate an arbitrary point on a segment. By choose a point on a line segment that has a certain relationship to other geometric figures.
Steps for Segment Addition Postulate:

Postulate: Segment Addition Postulate

Point B is a point on segment AC, that is the point B is between A and C, if and only if

AB + BC = AC

Segment Addition Postulate (SAP) states that if point B is between point A and point C, then AB + BC = AC. If AB + BC = AC, then B is between A and C.

segement postulate

In geometry, the segment addition postulate is defined as, if B is between A and C, then AB + BC = AC. The segment addition postulates are used in most of the proofs involving the congruence of segments.
Examples for Segment Addition Postulates:

Let's say that segment WV is congruent to segment VY and that segment XV is congruent to segment VZ. That is given, and we want to prove that segment XY is congruent to segment WZ.

The first step involves using the segment addition postulate. We can say that XV + VY = XY and WV + VZ = WZ both by the segment addition postulate.

The next step involves getting XY = VZ + WV by the substitution property. Since XY and WZ both equal WV + VZ, we can get XY = WZ by the substitution property. Thus, by the definition of congruent segment, the segment XY is congruent to segment WZ.

segement postulate

Example problems:

1) B is between A and C. Find the value of x and the measure of BC if: AB = 3, BC = 5x + 1, AC = 9.

Segement addition

Solution:

AC = AB + BC

9 = 3 + 5x + 1

9 = 5x +4

5x = 9 -4

5x =5

x =1 then

BC = 5(1) +1

BC = 6 (or)

AC =AB + BC

BC = AC - AB

= 9 - 3

BC =6

2) Z is between X and Y, find the value of x and the measure of YZ and XZ. XY = 24, YZ = 3x, XZ = 7x – 4.

Solution:

XY = XZ + YZ

XY = (7x + 4) + (3x)

Xy = 7x + 4 + 3x

XY = 10x + 4

24 = 10x -4

10x = 24 - 4

x =2

XZ = 7(2) + 4

XZ =14 + 4

XZ = 18

YZ = 3(2)

YZ = 6

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