They are equal
Red
Green

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As always, concise and elegant solution! I appreciate that you make the proof more readable by only stating the results of the angle chasing. It makes reading way more engaging as the reader is asked to quickly verify the statements for himself.

One litte question: The only reasoning I can see for y being the the length of the altitude of triangle BOC through C is the following: Let B' be the diametrically opposite of B. <COB' = 90° - <AOB. So, triangle ABO can be placed with B = O and O = C so that F is in line BB'. As <BFO = 90°, F is the the foot of the altitude and y the length of it.(Triangle COD and x is a similar case.)

What was your reasoning? Have you found a shorter explanation?

Keep up the good solutions!

Christian Helms
- 2 years ago

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Thanks! $y$ is the same length as the altitude of $\triangle BOC$ because $\angle COF = 90°$ and $\angle BFO = 90°$ . (If you drew a point $H$ such that $BH$ is the altitude of $\triangle BOC$ , then $BHOF$ would be a rectangle and therefore $y = FO = BH$ .)

David Vreken
- 2 years ago

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Oh right, I didn't consider that we can also take the altitude through be $B$ instead of $C$ because we also know $\overline{CO}$ and not only $\overline{OB}$ . This is actually fairly obvious, I should have given it a little more thought. My reasoning was overly complicated. Thanks for replying!

Christian Helms
- 2 years ago

How area's of ΔBOC and Δ ODE are equal man??

indian man
- 1 year ago

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For $\triangle BOC$ , take $OC = r$ as the base, and $OF = y$ as the height, so its area is half base times height or $\frac{1}{2}ry$ . For $\triangle ODE$ , take $OE = r$ as the base, and $DG = y$ as the height, so its area is half base times height or $\frac{1}{2}ry$ . Therefore, both $\triangle BOC$ and $\triangle ODE$ have areas that are equal to $\frac{1}{2}ry$ .

David Vreken
- 1 year ago

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Thanks ,it will be helpful if you resolve my another strange doubt,

Let intersection point of lines CD and OE be M in your diagram above. ∠OMC = ∠DMG (vertically opposite) and other angle is 90 degree so third angle of both should be same by angle sum property. hence Δ OMC ~ ΔDMG (is congruent to) by AAA .Now we know that in congruent triangles sides opposite to equal angles are equal ,Hence OC = DG or, y=r but as you can see y can never be equal to radius as it is in one half of circle and not drawn from radius. So I wonder whether I am wrong in my concepts or the diagram is not correct??

indian man
- 1 year ago

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@Indian Man – AAA is not a congruence property but a similarity property, so $\triangle OMC$ and $\triangle DMG$ are not congruent with congruent sides but similar with proportional sides. (Think about two differently sized equilateral triangles - they both have the same angles but are not congruent to each other.)

David Vreken
- 1 year ago

*
{pi/4 + sqrt(2)/2}.
The small green areas are each a sector with central angle of 45 degrees less a triangle with 2 sides = radius and an included angle of 45 degrees = r^2
*
{pi/4 - sqrt(2)/2}. Combining these terms, the green area is given by pi*r^2/2, which is 1/2 of the circle, so the areas are equal.

4 Helpful
3 Interesting
1 Brilliant
1 Confused

Do you have to prove AE is a diameter to know the angles for the small sectors?

Christoph Cox
- 2 years ago

'The small green areas are each a sector with central angle of $45^\circ$ ...'

This assumes that arc AB and arc DE are congruent. They have to add up to $90^\circ$ , but could be different. The proof would have to be modified to show that the green area is still $\frac {1}{2} \pi r^2$ in the general case.

Matthew Feig
- 2 years ago

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it could be taken as a wlog(without loss of generality) answer

MegaMoh .
- 9 months ago

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Let the center of the circle be $O$ and draw $AO$ , $BO$ , $CO$ , $DO$ , and $EO$ . Also draw a perpendicular to $AO$ from $B$ at $F$ and a perpendicular to $EO$ from $D$ at $G$ .

Since inscribed angles $\angle ABC = \angle BCD = \angle CDE = 45°$ , central angles $\angle AOC = \angle BOD = \angle COE = 90°$ . Therefore, $AE$ is a diameter. Also, by the angle sums of triangles it can be found that $\angle FBO = \angle GOD$ and $\angle FOB = \angle GDO$ , and since $BO = DO = r$ as radii of a circle, $\triangle FOB = \triangle GDO$ by ASA congruence. Therefore, $FO = GD = y$ and $BF = OG = x$ .

Both the areas of $\triangle BOC$ and $\triangle ODE$ are equal to $\frac{1}{2}ry$ , so they can be swapped and still preserve area; and both the areas of $\triangle COD$ and $\triangle ABO$ are equal to $\frac{1}{2}rx$ , so they can be swapped and still preserve area. Therefore, the green and red regions are both equivalent to areas of half a circle, so

they are equal.