Can you find Angle X in the Cyclic Quadrilateral? | Important Geometry skills explained
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- Learn how to find Angle X in the Cyclic Quadrilateral. Learn geometry skills: Isosceles Triangles property, angles at circumference and center. Step-by-step tutorial by PreMath.com
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Can you find Angle X in the Cyclic Quadrilateral? | Important Geometry skills explained
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Hey professor, great video👍 keep it up, good work!
Thanks, will do!
Glad you think so!
Thanks for your feedback! Cheers!
You are awesome, Medrain. Keep it up 👍
Love and prayers from the USA! 😀
My construction was a bit simpler. I joined B to C with a line segment and C to D with a line segment, thus forming two congruent isosceles triangles CDE and CBD. The angles at E and B then each had to be half the angle at D.
Congratulations Einstein.
excellent
Alternative step 4: construct a line segment connecting points C and D. Triangles CBD and CDE will be congruent based on SSS => angle CBD = angle CED = x.
Quadrilateral CBDE: Interior angles of a quadrilateral add up to 360° => 140° + 110° + 2x = 360°, 2x = 110° => x = 55°
Very informative video thanks for sharing
Different solution -
Measure of angle EAB = 70. Hence, measure of Central Angle ECB = 140.
∆CED is congruent to ∆DCB (by S-S-S congruence theorem).
Hence measure of angle ECD = measure of angle DCB = 70
Therefore measure of angle CED = x = measure of angle CDE = 55 (because ∆CED is isosceles and measure of angle ECD = 70).
Awesome👍
Thanks for sharing😊😊
Angle A=70º ⇒ Angle D=180º - 70º=110º.
CE=CD=CB and ED=BD ⇒ the triangles CED and CDB are isosceles and equal ⇒ The
angle D=2X ⇒ X=110º/2=55º
Amazing. Thank you.
Great video👍👍
Thanks for sharing😊😊
Since CB=CE and DB=DE, then
Very good, Osumanu dear
Excellent!
You are very welcome!
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You are awesome. Keep rocking 👍
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Stay blessed 😀
I found your method more simple and clear. Thank you for sharing.
@@Abby-hi4sf Thank you for acknowledging it.
Very good. You explain very professionally. Many thanks
Thanks for video. Good luck sir!!!!!!
Brilliant explanation. Really enjoyed it. 👍👍👍👍
Interesting and impressive. Very interesting
Once angle EDB was established to be 110 degrees, by joining radius CB, a symmetrical figure CBDE is produced, made up of 2 congruent triangles, and line CD would divide angle EDB into equal halves of 55 degrees.
Since CD is also a radius equal to CE, triangle CDE is isosceles, so angle CDE = angle X = 55 degrees.
Many thanks teacher
Basit bir yol olarak:
1. BDE yayı 70.2=140 derecedir.
2. ED yayı BD yayına eşit olduğu için ikisi de 70'er derece olur.
3. Çemberin merkezi yani C'den D'ye bir doğru çizersek yeni oluşan ECD açısı 70 derece olur.
4. EC ve CD yarıçap olduğu için birbirine eşittir. Yani CDE açısı da x olur.
5. 2x=180-70=110
x=55.
Ben olsam böyle çözerdim. Daha pratik.
5.
Quite an extravagant task! But maybe this situation will happen somewhere in real life.
Thank you for the opportunity to refresh my school knowledge of planimetry, Mr. PreMath. God bless you, your family and your wonderful country!
You are very welcome!
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شكرا شرح رائع
Pretty straightforward
hello sir, its me again!!!
Welcome back! Good to see you again.
You are awesome, Haofeng. Keep it up 👍
Since CE = CD = CB = radius, therefore the 2 isosceles triangles (CDE and CDB) are congruent (SSS rule). Also opposite angles (A and D) are supplementary in cyclic quadrilateral.
So, D = 180 - 70 = 110. Since angle at the center of equal chords (DE and DB) are equal (let's say Z). So x + x = 180 - Z = (110 - x) + (110 - x). So, 4x = 220. So x = 55.
wow amazing video.
also please can you do hard math olympiad problems to get ready for a math olympiad?
thanks a lot for your help
nice
Triangle ECD is identical to triangle BCD, so angle EDB=2X, therefore X=55 degrees. What a great problem.
Thnku
You are very welcome!
So nice of you, Pranav
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V nice explanation
Thanks for liking
Excellent!
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Stay blessed 😀
I have solve the sum mr Locke system.
angle ECB=140degree
therefore ang CED+ang CBD=360-(110+140)= 110 degree.
hence answer as triangle ECB and CBD are congruent to each other as S-S-S.
Easy 😎
By drawing CB and CD, you have two equal isosceles triangles CED and CDB.
Angles CED = CDE = CDB = CBD = x
Therefore ECD = DCB = ECB/2
ECB = 140 for stated reason,
therefore ECD = DCB = 70
ECD + CED +EDC = 180
70 + 2*CED = 180
CED = (180 - 70)/2 = 55
The problem became easier after seeing the smaller triangles formed
👍
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Stay blessed 😀
Could have connected CD, then by symmetry
Absolutely
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Central angle is 70 . 180-- 70 = 110 . So each angle is 55.
Your answer is fascinating but you could just draw another triangle and prove that triangle is congruent then you've 2x = 180-70 , x =55
Hi sir ! Completed 12th and joining bsc mathematics , Inspiration for 100s...
Good 👍 Keep rocking, Aathi
Thanks for sharing! Cheers!
You are awesome. Keep it up 👍
Love and prayers from the USA! 😀
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@@PreMath Thank You sir !
(180-70)÷2
55.
55°
Excellent!
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You are awesome, Dilshan. Keep it up 👍
asnwer=65 isit
Long procediment video
Better:
1° ang BDE = 2(70)=140
2° segmens equals ==> arcs equals
AE= BD ===> arcED = arcBD=70
3° angcenterECD = 70
4° traz radios CD and you have isos
===> X=35
Steps in 10segs, mental
Made in Peru
First solve yourself than look the solution, sometimes you end up same approach or your own very distinct approach to the problem.
ED=BD given
Draw CE = CD =CB= radius
From above,
triangle ECD & triangle ECB are equal ----1
Angle ECB = 2 (Angle EAB) = 140 ----------2
From 1 & 2,
Angle ECD = Angle DCB = 70 ------------------3
From 1,
Angle CED = Angle CDE = × -------------------4
From triangle ECD,
Angle ECD + Angle CED + Angle CDE =180
70 + × + × = 180
×= 110/2=55
Angle CED=1/2 Angle BDE
"Opposite angles in a cyclic quadrilateral add up to 180°"
Why is that? Is there any proof?
@@martinwestin4539 That proof is very good indeed and it is easy as well. Thank you!
75
2x = 110
X = 55
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Let's try this one before watching the video:
- Draw a straight line from E to B => α + δ = 180° => δ = 180° - α => δ = 180° - 70° => δ = 110° (opposite angles on both sides of a chord add up to 180°)
- Draw a straight line from C to D
- Since ED = BD => CD = bisector of the angle BDE
- Angle CDE = 55°
- CD = CE (= r)
- Triangle CDE is an isosceles triangle.
- Angle CDE = Angle CED
- Angle CED (x) = 55°
Solved mentally..
Before watching: angle ECB („at the center“) is double of angle EAB („at the circumference“) thus 140. The two triangles ECD and DCB are congruent (same circle sector length as one side and the other sides are the radius), thus they are also isosceles triangles. Since ECB is 140 degrees, ECD is half, thus equally 70 degrees. x half the remainder to 180, thus 110/2 or 55 degrees (due to the fact that the triangle is isosceles). So c is simply 55 degrees.
Now I watch!
After watching: ok I had the right solution, but I think my way was a bit more „direct“! 😀
Easy. It is at CED 😂
x = 55 degree in just Seconds
Your method is very lengthy. There is short method.