i need some help with a question on circle geometry
any help would be great

POints A,B,C,D,E,F lie on teh circumference of a circle
Prove that <ABC+<CDE+<EFA=360

thanx

2. Originally Posted by shosho
i need some help with a question on circle geometry
any help would be great

POints A,B,C,D,E,F lie on teh circumference of a circle
Prove that <ABC+<CDE+<EFA=360

thanx
Hello,

I've attached a drawing of the circle and the 6 points.

You are supposed to know that an angle at the perimeter of a circle is half as large as the angle at the centre both subtended over the same arc.

I've marked corresponding angles with the same colour.

Now consider the angles at the centre of the circle: The black angle + the thick blue angle = 360° and the red angle + the thin blue angle = 360°. Together they have 720°. The sum of angles at the perimeter have the half of 720° thus 360°

3. Hello, shosho!

Points $A,B,C,D,E,F$ lie on the circumference of a circle.
Prove that: $\,\angle ABC +\angle CDE+\angle EFA\:=\:360^o$
An inscribed angle is measure by one-half its intercepted arc.

$\angle ABC \:=\:\frac{1}{2}\left(\overline{CD{E}FA}\right) \:=\:\frac{1}{2}\left(\overline{CD} + \overline{DE} + \overline{EF} + \overline{FA}\right)$

$\angle CDE \:=\:\frac{1}{2}\left(\overline{EFABC}\right) \:=\:\frac{1}{2}\left(\overline{EF} + \overline{FA} + \overline{AB} + \overline{BC}\right)$

$\angle EFA \:=\:\frac{1}{2}\left(\overline{ABCDE}\right) \:=\:\frac{1}{2}\left(\overline{AB} + \overline{BC} + \overline{CD} + \overline{DE}\right)$

Then: . $\angle ABC + \angle CDE + \angle EFA$

. . $= \:\frac{1}{2}\left(\overline{CD} + \overline{DE} + \overline{FA}\right) \;+ \;\frac{1}{2}\left(\overline{EF} + \overline{FA} + \overline{AB} + \overline{BC}\right) \;+ \;\frac{1}{2}\left(\overline{AB} + \overline{BC} + \overline{CD} + \overline{DE}\right)$

. . $= \:\frac{1}{2}\left(2\!\cdot\!\overline{AB} \:+ \: 2\!\cdot\!\overline{BC} \:+ \: 2\!\cdot\!\overline{CD} \:+ \:2\!\cdot\!\overline{DE} \:+ \:2\!\cdot\!\overline{EF} \:+ \:2\!\cdot\!\overline{FA}\right)$

. . $= \:\overline{AB} + \overline{BC} + \overline{CD} + \overline{DE} + \overline{EF} + \overline{FA}$

. . $= \;360^o$

4. could i solve this question by splitting the original cycliv quad into 2 cyclic quads?

but my question is whether a cyclec quad is any quadrilateral with 4 points on teh circle or if it has to overlap the centre as well?

5. Originally Posted by shosho
i need some help with a question on circle geometry
any help would be great

POints A,B,C,D,E,F lie on teh circumference of a circle
Prove that <ABC+<CDE+<EFA=360

thanx
Now I'm prepared to bet that this is in fact not true (note under-statement here).

For it to be true you have probably either missed a condition or

RonL

6. Originally Posted by shosho
i need some help with a question on circle geometry
any help would be great

POints A,B,C,D,E,F lie on teh circumference of a circle
Prove that <ABC+<CDE+<EFA=360

thanx
For a solution to the question probably intended rather than the one asked see here. You should be able to concoct a really neat proof from what is there. (The proof there is so neat it has been accepted for publication)

RonL