Calculate the angle X and justify | Learn how to Solve this Tricky Geometry problem Quickly

Learn how to find the angle X in the given diagram. Solve this tricky geometry problem by using Cyclic Quadrilateral and alternate interior angles Theorems.

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Calculate the angle X and justify | Learn how to Solve this Tricky Geometry problem Quickly

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Olympiad Mathematics
pre math
Po Shen Loh
Learn how to find the angle X
How to Solve this Tricky Geometry problem Quickly
Cyclic Quadrilateral Theorem
Alternate interior angles
premath
premaths

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[timeonly]

Started same as you...got angle ADC = 40 degrees, which is an inscribed angle that subtends minor arc APC (= 2 x 40 = 80 deg). Then, major arc ABCD = 360 - 80 = 280 deg. But angle APD is an inscribed angle for that major arc, and half that measue, so x = 1/2 x 280 = 140 deg. Done!

[larisamedovaya]

It’s easy. 180-140=40 ; two parallel lines and equal angles. Arc corresponds to angle 40 Degree is 80 Degree. Whole circle has 360 Degree. Angle “x” stays on 360- 80= 280 Degree arc. Inscribe in circle “x” is half of the arc 280 Degree; Therefore angle “x” =140 Degree .GED

[montynorth]

Point P is not required to be at one fixed position on arc AC.
If point P is positioned such that PC is parallel to AD, then a symmetrical trapezoid is formed, APCD, with angles 40, 40, 140, 140.

[alexundre]

Good night Master
Congratulations on the Class
Big hug from Brazil 🇧🇷
Thank you

[HappyFamilyOnline]

Great explanation👍
Thanks for sharing✨✨

[APUS_NUNN]

Sehr schöne Aufgabe. Ebenso schöne Lösung. Dankeschön.

[jhill]

Learn a new, obscure geometry rule quite often.

[Waldlaeufer]

DAB = ADC = 40°
Therefore, CPA = 180° - 40° = 140°

[santiagoarosam]

The angle ∠APC=X subtends the chord AC → The vertex P can be at any interior point of the arc AC and will continue to be ∠APC=X → If we place the point P at the outer end of the perpendicular radius of AC, the premises do not change of the problem and the central angle ∠AOC is bisected by ∠AOP and ∠POC → If AB and CD are parallel → ∠BAD=∠ADC=180º-140º=40º → ∠ADC subtends AC → The corresponding central angle ∠AOC=2* 40º=80º → ∠AOP=∠POC=80º/2=40º → ∆POC=∆AOP are isosceles → ∠PAO=∠APO=∠ CPO=∠ PCO=(180º - 40º)/2=70º → ∠APC=∠ APO+∠CPO=2*70º=140º → X=140º
Thanks and greetings to all

[mathpuzzles]

Nice solve! you reminded of theorems! Thanks for making it!