Two Methods to find the Angle X in the Cyclic Quadrilateral | Important Geometry skills explained

Today I will teach you tips and tricks to solve the given olympiad math question in a simple and easy way. Learn how to prepare for Math Olympiad fast!

Need help with solving this Math Olympiad Question? You're in the right place!

I have over 20 years of experience teaching Mathematics at American schools, colleges, and universities. Learn more about me at

Two Methods to find the Angle X in the Cyclic Quadrilateral | Important Geometry skills explained

Olympiad Mathematical Question! | Learn Tips how to solve Olympiad Question without hassle and anxiety!

#FindAngleX #GeometryMath #MathOlympiad
#OlympiadMathematicalQuestion #HowToSolveOlympiadQuestion #MathOlympiadQuestion #MathOlympiadQuestions #OlympiadQuestion #Olympiad #AlgebraReview #Algebra #Mathematics #Math #Maths #MathOlympiad #HarvardAdmissionQuestion #Circle #Circles
#MathOlympiadPreparation #LearntipstosolveOlympiadMathQuestionfast #OlympiadMathematicsCompetition #MathOlympics #CyclicQuadrilateral #IsoscelesTriangles #FindAngleX #CollegeEntranceExam #CollegeEntranceQuestion
#blackpenredpen #ExponentialEquations #SolveExponentialEquation #MathOlympiadTraining #Olympiad Question #GeometrySkills #GeometryFormulas #AreaOfTriangles #AreaOfRectangle
#MathematicalOlympiad #OlympiadMathematics

How to solve Olympiad Mathematical Question
How to prepare for Math Olympiad
How to Solve Olympiad Question
How to Solve international math olympiad questions
international math olympiad questions and solutions
international math olympiad questions and answers
olympiad mathematics competition
blackpenredpen
math olympics
olympiad exam
olympiad exam sample papers
math olympiad sample questions
math olympiada
British Math Olympiad
olympics math
olympics mathematics
olympics math activities
olympics math competition
Math Olympiad Training
How to win the International Math Olympiad | Po-Shen Loh and Lex Fridman
Po-Shen Loh and Lex Fridman
Number Theory
There is a ridiculously easy way to solve this Olympiad qualifier problem
This U.S. Olympiad Coach Has a Unique Approach to Math
The Map of Mathematics
mathcounts
math at work
exponential equation
system of equations
solve system of equations
solve the equation
Solve the Logarithmic Equation
Pre Math
Olympiad Mathematics
Olympiad Question
Cyclic triangles
Isosceles triangles
Find angle X
Geometry
Geometry math
Geometry skills
Right triangles
College Entrance Exam
College Entrance Question

Subscribe Now as the ultimate shots of Math doses are on their way to fill your minds with the knowledge and wisdom once again.

[sameerqureshi-khcc]

Sir I m math teacher and after subscribing your channel I realized that math is not as difficult as many people consider, love from Pakistan 🇵🇰

[nineko]

There is a third method, once you know that angle AOC is 92, you can already say that angle ABC is AOC/2.

[dannymeslier]

Yet another way to look at it:
Draw line BD. BD bisects x because chords AD and DC are equal. By Thales theorem, <ADB = 90°, so x/2= 180 - 90 - 67 hence x = 46.

[philippeganty]

A bit faster: Triangles OAD and ODC are isosceles and equal, and triangle OCB is isosceles. Therefore, the angles OAD + ODA + ODC +OCD = 4*67°, and the angles OCB = OBC = X. The sum of the angles of the quadrilateral ABCD = 360°, therefore, 360° = 4*67°+ 2*X → X = 46°.

[Ensign_Cthulhu]

2:50 I proved that angle ODC had to equal angle AOD because triangles AOD and COD were congruent side-side-side, thus inadvertently proving the angles subtended by the same chord length theorem (which I had forgotten!), and then continued with the angles on a straight line.

[manojitmaity]

You may use the exterior angle theorem when you add the centre 'o' with the vertex 'c'.

Thus you can conclude <AOC= <OCB +<OBC=X+X =2X

[BesseDenmark]

Or we can use cyclic quadrilateral theorem to calculate angle DCB. It's 113 degrees . Now let's draw a segment from A to C. We divided the quadrilateral into two triangles. Triangle ABC is a right triangle because one of its sides is at the same time the diameter of the circle. So triangle ADC has two 23 degrees angles and one 134 degrees angle. Now we can use quadrilateral theorem again. It is 134 degrees + x = 180 degrees. Therefore x is equal to 46 degrees.

[tombufford]

At a quick glance: 4 sided figure then interior angles sum to 360. OAD and ODC are equal isosceles triangles then ODA, ODC and OCD angles = 67. Angle AOD =180-(2 *67) = 46 = Angle DOC. Then angle COB = 180 - 92 = 88. OCB is isosceles 2*x= 180-88. x= 46 degrees.

[ujwalsmanhas]

tri OAD and tri ODC are congruent. radius and equal side so <adc = 67*2 .so <x =180-67*2 = 46 .

[kennethstevenson]

The four sided figure is made up of three isosceles triangles because the equal sides are radii. This enables a transfer of information to the last triangle which has two equal angles who's sum is 180-88 = 92. 92/2=46 x=46 Solution by diagram. Great!

[zplusacademy]

Extremely beautiful ❤️🙏 sir as always ❤️🙏🙏🙏🙏

[alexundre]

Good Morning Master
Thank you Very much for the instruction
A Hug From in Brasil 🇧🇷
Grateful

[rbdgr]

Join BD. In rt triangle ADB, angle ABD = 23°. AD and CD subtends equal angles at B. Hence, x = 2*angleABD = 2*23° = 46°.

[DDX]

I have solved it by second methed. But now I know the 1st method too thanks my HERO

[harikatragadda]

ABCD is a Cyclic Quadrilateral, with opposite corner angles adding to 180°.
X + 2*67 = 180
X = 46°

[egillandersson]

By the first method, once you have established that angle AOC is 2x46*, you can conclude because it's the central angle that intercepts the same chord as the inscribed angle ABC. So this latter is half AOC. Sorry for my bad english !

[trishanuagarwal]

Sir can u please xplain how u did COD=46°

[MarieAnne.]

Another method using central angles and angles subtended by an arc on the circle, using the following theorem:
_"The angle subtended by an arc of a circle at its center is twice the angle it subtends on the circle’s circumference."_

First we calculate central angle DOB using theorem above:
∠DOB = 2 × ∠DAB = 2 x 67° = 134°
Next we calculate central angle AOD using the fact straight angle AOB = 180°
∠AOD = 180° - ∠DOB = 180° - 134° = 46°
Since chords AD and DC are congruent, the so are the central angles they subtend:
∠DOC = ∠AOD = 46°
∠AOC = ∠AOD + ∠DOC = 46° + 46° = 92°
Finally, we calculate angle ABC subtended by arc AC using central angle AOC (using theorem above):
*∠ABC = x = 1/2 × ∠AOC = 1/2 × 92° = 46°*

[OldSloGuy]

A shorter solution: Erect a tangent (T) at A, The angle TAD = 90 - 67 = 23, The angle DAC subtends a similar chord and thus is equal to TAD. The angle ACB = 90, The rest is arithmetic.

[barttemolder]

ACB is always a right angle.
DO is perpendicular to AC so it is parallel to CB.
Thus angle AOD is the same as OBC.
You only need to calculate AOD. X is AOD.