©2023 Saltire Software ← GXWeb Polygon Playground 1 → GXWeb Polygon Playground 2: Angle Theorems

  

GXWeb Polygon Playground 2:
Angle Theorems

Saltire Software, home of Geometry Expressions and GXWeb

GXWeb Polygon Playground 1

GXWeb Polygon Playground 2: YouTube Introduction (5:37)

Introduction Hints to Get Started GXWeb Polygon Model Build Your Own GXWeb Model Mathematical ToolKit Share Your Results Behind the Scenes

Try the Full GXWeb Construction

 

---

---

 

Show Intro

Introduction

Back to Top

Jump to Model

Here is a collection of geometry puzzles.

GXWeb has been enhanced to generate an abundance of relationships among the angles of cyclic polygons, and you are challenged to try to solve these.

Use the New Problem button to explore new polygonal challenges and try to solve each - perhaps by building your own model using GXWeb!

Configure your questions: proof, numeric, algebraic or random. Score points for correct answers to algebraic and numeric theorems, and try your hand at the proofs - then share your results with others.

Generate PDF examples from the Examples Collection - always looking for patterns and relationships because there lies the heart of mathematics.

Hide Intro

Back to Top

Hints to Get Started

Jump to Model

Build Your Own GXWeb Model

Mathematical ToolKit

Share Your Results

 

---

---

 

Show Hints

Some Hints to Get Started

Back to Top

Jump to Model

And now for some hints to help you get started...

---

Tap each of the images following to see how the embedded GXWeb applet on this page can offer a quick and easy live model for the generated challenge problems, and assist you in solving these.

---

 

Show  In addition to the angles of a triangle adding up to 180 degrees, you'll need to use the facts that opposite angles of a cyclic quadrilateral add up to 180 degrees and that two angles at the circumference which sit on the same chord are equal.

For example, in Fig. 1, ACDE is a cyclic quadrilateral, hence angles $CAE=180-z$. But ABCE is also a cyclic quadrilateral, hence $BCE+BAE=180$. As BAE=BAC+CAE, we have $x+y+(180-z)=180 => z = x+y$. In the GXWeb model of Fig. 1, BDEF is a cyclic quadrilateral, hence angles $DBF=180-DEF$. But BCDF is also a cyclic quadrilateral, hence $CDF+CBF=180$. As CBF=CBD+DBF, we have $DEF=x+y$.	⇑ Tap to see this in GXWeb ⇓ Try the Full GXWeb Fig 1 Construction
In Fig. 2, DFE and DCE both stand on the chord DE, and hence have the same value. So $DCE=x$. Hence $DCB=x+y$. But DCBA is a cyclic quadrilateral so $x+y+z=180$. In the GXWeb model of Fig 2, EGF and EDF both stand on the chord EF, and hence have the same value. So $EDF=x$. Hence $EDC=x+y$. But EDCB is a cyclic quadrilateral so $x+y+z=180$ (where $CBE = z$).	⇑ Tap to see this in GXWeb ⇓ Try the Full GXWeb Fig 2 Construction

---

 

Show  For problems which contain lines through the circle center, you may need the additional result that the angle on a chord at the center of the circle is twice the angle on that chord at the circumference.

In Fig. 3, if angle OEA is $x$, then so is angle OAE, as the triangle is isosceles. So angle $AOE = 180-2x$. Hence $ADE=90-x$. If $ADE=y$, then $CDE=90-x+y$. But BCDE is a cyclic quadrilateral, so $CDE+CBE=180$. If $CBE=z$, then we have $90-x+y+z=180$, or $y+z=x+90$. In this GXWeb model, if angle AFB is $x$, then so is angle ABF, as the triangle is isosceles. So angle $BAF = 180-2x$. Hence $BEF=90-x$. If $DCF=y$, then $DEF=90-x+y$. But CDEF is a cyclic quadrilateral, so $DEF+DCF=180$. If $BED=z$, then we have $90-x+y+z=180$, or $y+z=x+90$.

⇑ Tap to see this in GXWeb ⇓

---

 

Show  For figures which have angles which do not lie on the circumference of the circle, the result in figure 4 may be useful. The angle between chords AC and BD, which intersect outside the circle is half the difference between the angles AOB and COD. The angle between the chords AD and BC, which intersect inside the circle is half their sum.

In Fig. 4, if angle AOB is $x$ and angle COD is $y$, then angle $AEB =\frac{x}{2}-\frac{y}{2}$ AND angle $AFB =\frac{x}{2}+\frac{y}{2}$. In this GXWeb model, if angle BAD is $x$ and angle CAE is $y$, then angle $CGE =\frac{x}{2}-\frac{y}{2}$ AND angle $BFD =\frac{x}{2}+\frac{y}{2}$.	⇑ Tap to see this in GXWeb ⇓
In Fig. 5, for example, $AOF=2\cdot ABF$ and $COE=2\cdot CAE$, which is $2\cdot (180-CDE)$. By fig. 4, $AGF=\frac{COE}{2}-\frac{AOF}{2}$, which is $180-CDE-ABF$. So $x+y+z=180$ In this GXWeb model, $BAG=2\cdot BCG$ and $DAF=2\cdot DBF$, which is $2\cdot (180-DEF)$. By fig. 4, $BHG=\frac{DAF}{2}-\frac{BAG}{2}$, which is $180-DEF-BCG$. So $x+y+z=180$	⇑ Tap to see this in GXWeb ⇓

---

 

Show  Where an angle is given between a line passing through the circle center and a chord, you may find this result useful (fig. 6).

In Fig. 6, given triangle ABC whose vertices lie on a circle centered at O, let the line joining A and O cut BC at D. Then $ADB=90+ACB-ABC$. In this GXWeb model, given triangle BCD whose vertices lie on a circle centered at A, let the line joining B and A cut CD at E. Then $BEC=90+BDC-BCD$.	⇑ Tap to see this in GXWeb ⇓
In Fig. 7, $DBE=z$ as it sits on the same chord as DAE. $BDE=x$ as it sits on the same chord as BCE. Hence $DFE=90-x+y$, or $x+z=90-y$. In this GXWeb model, $ECF=z$ as it sits on the same chord as EBF. $CEF=x$ as it sits on the same chord as CDF. Hence $EGF=90-x+y$, or $x+z=90-y$.	⇑ Tap to see this in GXWeb ⇓

 

---

Hide Hints

Back to Top

Hints to Get Started

Jump to Model

Build Your Own GXWeb Model

Mathematical ToolKit

Share Your Results

 

---

---

 

GXWeb Cyclic Polygon Problem Generator

This document requires an HTML5-compliant browser.   Let ABCDE be a cyclic pentagon with center O. Let G be the intersection of BC and ED. Prove that BAE+CGD+CAD = 180 (difficulty 14) Answer New Problem Proof How might you prove that BAE+CGD+CAD = 180? Let BAE=x. Let CGD=y. Let CAD=z. Let DCG=w. As DCG=w, DCB=180-w. As BCDA is a cyclic quadrilateral, BAD=180-BCD, so BAD=w. As BAD=w, BAC=w-z. As CGD=y, CDG=180-y-w. As CDG=180-y-w, CDE=y+w. As CDEA is a cyclic quadrilateral, CAE=180-CDE, so CAE=180-y-w. As CAE=180-y-w, CAB=x+y+w-180. But BAC=w-z, so x+y+w-180=w-z, or x+y+z=180, or BAE+CGD+CAD=180.

Problem Settings   Pentagon   Hexagon   Hexagon (four angles)   Heptagon   Octagon   Use Chord Intersections Problem type Proof problem Determine angle (algebraic) Determine angle (numeric) Random Create Collection Add to Collection Answer New Problem     Collection Settings Number of Examples: File Name   Shuffle   Introduction   Answers   Identifiers   Solutions solution for one in Running header       Custom Collection You can cut and paste identifiers of specific examples from the pdf's and put them here to build a custom pdf Identifiers (comma seperated) Custom PDF Clear List

Back to Top

Hints to Get Started

Jump to Model

Build Your Own GXWeb Model

Mathematical ToolKit

Share Your Results

 

---

---

 

Show GXWeb

Construct your own Model with GXWeb

Back to Top

Jump to Model

Back to Top

Hints to Get Started

Jump to Model

Build Your Own GXWeb Model

Mathematical ToolKit

Share Your Results

Hide GXWeb

 

---

---

 

Show ToolKit

Mathematical Toolkit

Back to Top

Jump to Model

Hide ToolKit