Finding an Angle using Cyclic Quadrilateral and Circle Theorems

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enter image description here
To find angle BGE from the diagram above, a proof was proposed:



"Angle BCF equals 100 degrees (external to triangle ACB);
ADEC and DBFC are cyclic quadrilaterals -> angle ADE equals 100 degrees; angle DEB equals 70 degrees.
Let´s call angle CBG alpha. Then, angle DCA equals 30+alpha and angle DCB equals 50-alpha (30+alpha+100+DCB=180 -> DCB=50-alpha). But because DECA is cyclic, angle DAE equals angle DCE.



Then angle BEG equals 80-alpha (external to triangle ABE). Now, we have:



80-alpha+alpha+BGE=180 -> BGE= 100 degrees."



Regarding the proof, I found the same thing up until the alpha part, which I found very confusing (I do not know why some angles equal what the proof has stated them to be). Hence, I would really appreciate if someone could find a proof excluding alpha and just involving theorems.



This is an alternate proof I had written up, but I'm still stuck on what to do next:
enter image description here



Help would really be appreciated. Thanks :)




geometry angle quadrilateral




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  • 1




    Just try to prove that CFGE is cyclic will do via angles ADC, AEC and AFB. Reasons to be quoted are "angles in the same segment" and "exterior angle of cyclic quadrilateral".
    – Mick
    Jul 20 at 7:33















up vote
0
down vote

favorite












enter image description here
To find angle BGE from the diagram above, a proof was proposed:



"Angle BCF equals 100 degrees (external to triangle ACB);
ADEC and DBFC are cyclic quadrilaterals -> angle ADE equals 100 degrees; angle DEB equals 70 degrees.
Let´s call angle CBG alpha. Then, angle DCA equals 30+alpha and angle DCB equals 50-alpha (30+alpha+100+DCB=180 -> DCB=50-alpha). But because DECA is cyclic, angle DAE equals angle DCE.



Then angle BEG equals 80-alpha (external to triangle ABE). Now, we have:



80-alpha+alpha+BGE=180 -> BGE= 100 degrees."



Regarding the proof, I found the same thing up until the alpha part, which I found very confusing (I do not know why some angles equal what the proof has stated them to be). Hence, I would really appreciate if someone could find a proof excluding alpha and just involving theorems.



This is an alternate proof I had written up, but I'm still stuck on what to do next:
enter image description here



Help would really be appreciated. Thanks :)




geometry angle quadrilateral




share|cite|improve this question















  • 1




    Just try to prove that CFGE is cyclic will do via angles ADC, AEC and AFB. Reasons to be quoted are "angles in the same segment" and "exterior angle of cyclic quadrilateral".
    – Mick
    Jul 20 at 7:33













up vote
0
down vote

favorite









up vote
0
down vote

favorite











enter image description here
To find angle BGE from the diagram above, a proof was proposed:



"Angle BCF equals 100 degrees (external to triangle ACB);
ADEC and DBFC are cyclic quadrilaterals -> angle ADE equals 100 degrees; angle DEB equals 70 degrees.
Let´s call angle CBG alpha. Then, angle DCA equals 30+alpha and angle DCB equals 50-alpha (30+alpha+100+DCB=180 -> DCB=50-alpha). But because DECA is cyclic, angle DAE equals angle DCE.



Then angle BEG equals 80-alpha (external to triangle ABE). Now, we have:



80-alpha+alpha+BGE=180 -> BGE= 100 degrees."



Regarding the proof, I found the same thing up until the alpha part, which I found very confusing (I do not know why some angles equal what the proof has stated them to be). Hence, I would really appreciate if someone could find a proof excluding alpha and just involving theorems.



This is an alternate proof I had written up, but I'm still stuck on what to do next:
enter image description here



Help would really be appreciated. Thanks :)




geometry angle quadrilateral




share|cite|improve this question











enter image description here
To find angle BGE from the diagram above, a proof was proposed:



"Angle BCF equals 100 degrees (external to triangle ACB);
ADEC and DBFC are cyclic quadrilaterals -> angle ADE equals 100 degrees; angle DEB equals 70 degrees.
Let´s call angle CBG alpha. Then, angle DCA equals 30+alpha and angle DCB equals 50-alpha (30+alpha+100+DCB=180 -> DCB=50-alpha). But because DECA is cyclic, angle DAE equals angle DCE.



Then angle BEG equals 80-alpha (external to triangle ABE). Now, we have:



80-alpha+alpha+BGE=180 -> BGE= 100 degrees."



Regarding the proof, I found the same thing up until the alpha part, which I found very confusing (I do not know why some angles equal what the proof has stated them to be). Hence, I would really appreciate if someone could find a proof excluding alpha and just involving theorems.



This is an alternate proof I had written up, but I'm still stuck on what to do next:
enter image description here



Help would really be appreciated. Thanks :)





geometry angle quadrilateral





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Jul 20 at 7:16









Cameron Choi

594




594







  • 1




    Just try to prove that CFGE is cyclic will do via angles ADC, AEC and AFB. Reasons to be quoted are "angles in the same segment" and "exterior angle of cyclic quadrilateral".
    – Mick
    Jul 20 at 7:33













  • 1




    Just try to prove that CFGE is cyclic will do via angles ADC, AEC and AFB. Reasons to be quoted are "angles in the same segment" and "exterior angle of cyclic quadrilateral".
    – Mick
    Jul 20 at 7:33








1




1




Just try to prove that CFGE is cyclic will do via angles ADC, AEC and AFB. Reasons to be quoted are "angles in the same segment" and "exterior angle of cyclic quadrilateral".
– Mick
Jul 20 at 7:33





Just try to prove that CFGE is cyclic will do via angles ADC, AEC and AFB. Reasons to be quoted are "angles in the same segment" and "exterior angle of cyclic quadrilateral".
– Mick
Jul 20 at 7:33











1 Answer
1






active

oldest

votes

















up vote
1
down vote



accepted










Here is the picture with hints.



enter image description here



Added:



All angles marked in red are equal (somehow). [By considering the large circle, the leftmost red is equal to rightmost red. By considering the small circle, the leftmost red is equal to the middle red. Note that their actual values need not be found because they will be equal definitely.]



From the above, we can say the bottom right quadrilateral is cyclic because of "exterior angle equals to the interior opposite angle". Since that quadrilateral is cyclic now, we can say that the green marked angle is $100^0$.






share|cite|improve this answer























  • Am I meant to find the values of the red angles? Thank you for the hint, but I am still a little confused on how to use them...
    – Cameron Choi
    Jul 20 at 8:07










  • See added in the hint.....
    – Mick
    Jul 20 at 18:25










  • Oh okay so the leftmost red angle is equal to the middle red angle because they both subtend arc A. And then x+(the middle red)=180, meaning that the bottom right quadrilateral is cyclic as the opposite angles (angle CEG and angle CFG) are supplementary. Then I use "exterior angle equals to the interior opposite angle" to find angle BGE, right? And does the leftmost red equal the rightmost red as the leftmost red is an exterior angle to the cyclic quadrilateral BDCF? Also, thank you!
    – Cameron Choi
    Jul 21 at 1:50







  • 1




    You've got the point. Note that using "exterior angle equals to the interior opposite angle" is one step faster and more direct than hiring a supplementary angle as agent.
    – Mick
    Jul 21 at 10:55










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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










Here is the picture with hints.



enter image description here



Added:



All angles marked in red are equal (somehow). [By considering the large circle, the leftmost red is equal to rightmost red. By considering the small circle, the leftmost red is equal to the middle red. Note that their actual values need not be found because they will be equal definitely.]



From the above, we can say the bottom right quadrilateral is cyclic because of "exterior angle equals to the interior opposite angle". Since that quadrilateral is cyclic now, we can say that the green marked angle is $100^0$.






share|cite|improve this answer























  • Am I meant to find the values of the red angles? Thank you for the hint, but I am still a little confused on how to use them...
    – Cameron Choi
    Jul 20 at 8:07










  • See added in the hint.....
    – Mick
    Jul 20 at 18:25










  • Oh okay so the leftmost red angle is equal to the middle red angle because they both subtend arc A. And then x+(the middle red)=180, meaning that the bottom right quadrilateral is cyclic as the opposite angles (angle CEG and angle CFG) are supplementary. Then I use "exterior angle equals to the interior opposite angle" to find angle BGE, right? And does the leftmost red equal the rightmost red as the leftmost red is an exterior angle to the cyclic quadrilateral BDCF? Also, thank you!
    – Cameron Choi
    Jul 21 at 1:50







  • 1




    You've got the point. Note that using "exterior angle equals to the interior opposite angle" is one step faster and more direct than hiring a supplementary angle as agent.
    – Mick
    Jul 21 at 10:55














up vote
1
down vote



accepted










Here is the picture with hints.



enter image description here



Added:



All angles marked in red are equal (somehow). [By considering the large circle, the leftmost red is equal to rightmost red. By considering the small circle, the leftmost red is equal to the middle red. Note that their actual values need not be found because they will be equal definitely.]



From the above, we can say the bottom right quadrilateral is cyclic because of "exterior angle equals to the interior opposite angle". Since that quadrilateral is cyclic now, we can say that the green marked angle is $100^0$.






share|cite|improve this answer























  • Am I meant to find the values of the red angles? Thank you for the hint, but I am still a little confused on how to use them...
    – Cameron Choi
    Jul 20 at 8:07










  • See added in the hint.....
    – Mick
    Jul 20 at 18:25










  • Oh okay so the leftmost red angle is equal to the middle red angle because they both subtend arc A. And then x+(the middle red)=180, meaning that the bottom right quadrilateral is cyclic as the opposite angles (angle CEG and angle CFG) are supplementary. Then I use "exterior angle equals to the interior opposite angle" to find angle BGE, right? And does the leftmost red equal the rightmost red as the leftmost red is an exterior angle to the cyclic quadrilateral BDCF? Also, thank you!
    – Cameron Choi
    Jul 21 at 1:50







  • 1




    You've got the point. Note that using "exterior angle equals to the interior opposite angle" is one step faster and more direct than hiring a supplementary angle as agent.
    – Mick
    Jul 21 at 10:55












up vote
1
down vote



accepted







up vote
1
down vote



accepted






Here is the picture with hints.



enter image description here



Added:



All angles marked in red are equal (somehow). [By considering the large circle, the leftmost red is equal to rightmost red. By considering the small circle, the leftmost red is equal to the middle red. Note that their actual values need not be found because they will be equal definitely.]



From the above, we can say the bottom right quadrilateral is cyclic because of "exterior angle equals to the interior opposite angle". Since that quadrilateral is cyclic now, we can say that the green marked angle is $100^0$.






share|cite|improve this answer















Here is the picture with hints.



enter image description here



Added:



All angles marked in red are equal (somehow). [By considering the large circle, the leftmost red is equal to rightmost red. By considering the small circle, the leftmost red is equal to the middle red. Note that their actual values need not be found because they will be equal definitely.]



From the above, we can say the bottom right quadrilateral is cyclic because of "exterior angle equals to the interior opposite angle". Since that quadrilateral is cyclic now, we can say that the green marked angle is $100^0$.







share|cite|improve this answer















share|cite|improve this answer



share|cite|improve this answer








edited Jul 20 at 18:34


























answered Jul 20 at 7:42









Mick

11.5k21540




11.5k21540











  • Am I meant to find the values of the red angles? Thank you for the hint, but I am still a little confused on how to use them...
    – Cameron Choi
    Jul 20 at 8:07










  • See added in the hint.....
    – Mick
    Jul 20 at 18:25










  • Oh okay so the leftmost red angle is equal to the middle red angle because they both subtend arc A. And then x+(the middle red)=180, meaning that the bottom right quadrilateral is cyclic as the opposite angles (angle CEG and angle CFG) are supplementary. Then I use "exterior angle equals to the interior opposite angle" to find angle BGE, right? And does the leftmost red equal the rightmost red as the leftmost red is an exterior angle to the cyclic quadrilateral BDCF? Also, thank you!
    – Cameron Choi
    Jul 21 at 1:50







  • 1




    You've got the point. Note that using "exterior angle equals to the interior opposite angle" is one step faster and more direct than hiring a supplementary angle as agent.
    – Mick
    Jul 21 at 10:55
















  • Am I meant to find the values of the red angles? Thank you for the hint, but I am still a little confused on how to use them...
    – Cameron Choi
    Jul 20 at 8:07










  • See added in the hint.....
    – Mick
    Jul 20 at 18:25










  • Oh okay so the leftmost red angle is equal to the middle red angle because they both subtend arc A. And then x+(the middle red)=180, meaning that the bottom right quadrilateral is cyclic as the opposite angles (angle CEG and angle CFG) are supplementary. Then I use "exterior angle equals to the interior opposite angle" to find angle BGE, right? And does the leftmost red equal the rightmost red as the leftmost red is an exterior angle to the cyclic quadrilateral BDCF? Also, thank you!
    – Cameron Choi
    Jul 21 at 1:50







  • 1




    You've got the point. Note that using "exterior angle equals to the interior opposite angle" is one step faster and more direct than hiring a supplementary angle as agent.
    – Mick
    Jul 21 at 10:55















Am I meant to find the values of the red angles? Thank you for the hint, but I am still a little confused on how to use them...
– Cameron Choi
Jul 20 at 8:07




Am I meant to find the values of the red angles? Thank you for the hint, but I am still a little confused on how to use them...
– Cameron Choi
Jul 20 at 8:07












See added in the hint.....
– Mick
Jul 20 at 18:25




See added in the hint.....
– Mick
Jul 20 at 18:25












Oh okay so the leftmost red angle is equal to the middle red angle because they both subtend arc A. And then x+(the middle red)=180, meaning that the bottom right quadrilateral is cyclic as the opposite angles (angle CEG and angle CFG) are supplementary. Then I use "exterior angle equals to the interior opposite angle" to find angle BGE, right? And does the leftmost red equal the rightmost red as the leftmost red is an exterior angle to the cyclic quadrilateral BDCF? Also, thank you!
– Cameron Choi
Jul 21 at 1:50





Oh okay so the leftmost red angle is equal to the middle red angle because they both subtend arc A. And then x+(the middle red)=180, meaning that the bottom right quadrilateral is cyclic as the opposite angles (angle CEG and angle CFG) are supplementary. Then I use "exterior angle equals to the interior opposite angle" to find angle BGE, right? And does the leftmost red equal the rightmost red as the leftmost red is an exterior angle to the cyclic quadrilateral BDCF? Also, thank you!
– Cameron Choi
Jul 21 at 1:50





1




1




You've got the point. Note that using "exterior angle equals to the interior opposite angle" is one step faster and more direct than hiring a supplementary angle as agent.
– Mick
Jul 21 at 10:55




You've got the point. Note that using "exterior angle equals to the interior opposite angle" is one step faster and more direct than hiring a supplementary angle as agent.
– Mick
Jul 21 at 10:55












 

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