Regular Polygon Inscribed in a Circle

• With centre O , draw the given circle.
• Draw a diameter AB and divide it into five equal parts (same number of parts as the number of sides) and number them as shown.
• With centres A and B and radius AB , draw arcs intersecting each other at M.
• Draw a line joining points M and A_2 intersecting the circle at I. Then AI is the length of the side of the pentagon.
• Starting from I , step – off on the circle, divisions IJ , JK , KL equal to AI.
• Draw segments AI , IJ , JK ,KL and LI thus completing the pentagon.

Continuous Curves of Circular Arcs

Let A , B , C , D and E are the given points.

• Draw line segments joining A with B , B with C , C with D and D with E.
• Draw perpendicular bisectors of AB and BC , intersecting each other at P.
• With P as centre and radius PA (or PB or PC) , construct an arc. This arc will pass through points A , B and C.
• Draw a line through P and C.
• Draw the perpendicular bisector of CD intersecting line through P and C at Q.
• With Q as centre , and radius QC (or QD) , construct an arc , passing through points C and D.
• Repeat the process. Please keep in mind that the centre of the arc is at the intersection of the perpendicular bisector and the line joining the previous centre with the last point of the previous arc.

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Trisecting a Right Angle

Let BAC be the give right angle.

• With centre A and any convenient radius , draw an arc cutting AB at Q and AC at P
• With the same radius and centres P and Q , draw arcs cutting the arc of step a) at R and S respectively. 
• Draw lines joining A with P and Q. AP and AQ trisect the right angle BAC.

                 Thus ∠BAR = ∠RAS = ∠SAC = 1/3 ∠BAC
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Division of Line Segment-Unequal Parts

Let AB be the given line segment to be divided into unequal parts say 1/6 ,1/5 ,1/ 4 ,1/3 ,1/ 2 .

• Draw a line segment AB of given length. 
• Draw perpendiculars AD and BC at points A and B. Complete the rectangle ABCD 
• Join diagonals AC and BD intersecting at E. 
• Draw perpendicular from E on AB and find intersection point F. 
• Now AF = 1/ 2 AB
•  Join D and F. The segment FD intersects the diagonal AC at G. Drop perpendicular from G to A. Then AH = 1/3 AB
• Similarly make constructions as given in the figure to obtain 1/ 4 AB , 1/5 AB and 1/6 AB

This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Perpendicular to a Line - 4

Following are the steps for construction of a perpendicular to a line from a point outside it. In this construction, the arcs are shown as circles. 

1. Let AB be the given line and P is the point outside it 
2. With A as centre and radius AP, draw an arc cutting the given line at C. 
3. With C as centre and radius CP, draw an arc cutting the arc of step 2 at R. 
4. Draw line segment joining PR intersecting AB at Q.

PQ is the required perpendicular. Drag Point P and see the effect. This is a Java Applet created using GeoGebra from www.geogebra.org - it looks like you don't have Java installed, please go to www.java.com

Perpendicular to a Line Segment - 3

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Perpendicular to Line Segment - 2

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Perpendicular to Line Segment from a point on it

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