circle theorems

[silk bloom]

X=38

[fossil marlin]

how

[eager sentinel]

<JHG = 52/2 = 26, since angles at the circumference are half that of at the centre

<OJG = (180-52)/2 = 64, since OJG is isosceles as 2 of its sides are radii, and base angles of isosceles are equal

<HGJ = 180-64 = 116, since <OJG and <HGJ are cointerior angles, as OJ || HG ( is parallel to)

hence x = 180 - 116 - 26 = 38, as angles in a triangle add to 180

[fossil marlin]

how do yk its talking abt angle jhg on the circumference

[eager sentinel]

since both angles are subtended by the same arc (JG)

or do you mean something else

[fossil marlin]

what does subtended mean

[eager sentinel]

like this

ig you can think of it if it forms a triangle with the same line of JG here

[wet elk]

i couldnt do that q myself 😭

[silk bloom]

It means made

[fossil marlin]

@eager sentinel can u help with other question

[eager sentinel]

im guessing you mean q10
i) well you know the C_2D and C_2B are both radii, so DC_2B is isosceles, so then x = 90 + y/2

ii)since angles at circumference are half that at the centre, you get that the reflex angle of <AC_1B = 2x, so the acute angle of <AC_1B is 360 - 2x = 180 - y
for the thing to be a cyclic quadrilateral, we need that <AC_1B + <AC_2B = 180
plugging in our values for the coords we get that

i guess it can be a bit tricky to visualise the step to get reflex of <AC_1B, so another way to think about it is by getting the cyclic quadrilateral PABD, so <APB = 180-x, and then as <APB and <AC_1B are both subtend by arc AB, you get that <AC_1B = 2 * <APB

[fossil marlin]

i dont get ii

[eager sentinel]

which part?

if you draw a point p in the drawing above, you get that APBD is a cyclic quadrilateral
as opposite angles in a cyclic quadrilateral add to 180 degrees, <APB + <ADB = 180, so <APB = 180 - x
as both <APB and <AC_1B are subtended by arc AB, <AC_1B = 2 * <APB, as angles at centre is double that at circumference when subtended by same arc, so <AC_1B = 360 - 2x = 360 - 2(90 + y/2) = 180 - y

for a quadrilateral to be cyclic (looking at AC_1BC_2 now), we require that its opposite angles must add to 180 degrees, so we need that <AC_1B + <AC_2B = 180, subbing in the values, we get that 180 - y + y = 180, so AC_1BC_2 is a cyclic quadrilateral

[fossil marlin]

i thought all vertices need to be on circumference

[eager sentinel]

we are looking at converse of that theorem now

so normally say if you have a circle and you draw 4 points in ityou get a cyclic quadrilateral

so then you know in that quadrilateral opposite angles add to 180 degrees

however

say if you take a quadrilateral

and its opposite angles happen to add to 180 degrees

you can then draw a circle on which all 4 points lie

so it is a cyclic quadrilateral

this is called taking the converse of theorems, for some theorems it doesnt work, but for all the circle theorems it does

[fossil marlin]

@upper delta