Relationship of circles and polygons

[Harry_the_cat]

Therefore angle x and Y are equal, because each triangle adds to 180

 

[Ihatdogs10]

okay I get it

 

[Harry_the_cat]

That's Angle, angle, angle AAA.
Therefore triangles are similar.

 

[Harry_the_cat]

OK. Now what do you know about the SIDES of simialr triangles?

 

[Ihatdogs10]

their sides?

 

[Harry_the_cat]

yes

 

[Harry_the_cat]

the sides of ANY similar triangles

 

[Ihatdogs10]

Side X and Y

 

[Harry_the_cat]

X is not a SIDE.


The ratio of corresponding sides are equal.
In the case here \(\displaystyle \frac{AB}{PQ}=\frac{AC}{PR}=\frac{BC}{QR}\).

 

[Ihatdogs10]

Am I right?

 

[Ihatdogs10]

Am I right?

ANGLE PX AND QY?

 

[Harry_the_cat]

Is that familiar to you?

 

[Harry_the_cat]

ANGLE PX AND QY?

PX and QY are SIDES not angles

 

[Ihatdogs10]

Is that familiar to you?

ahh right

 

[Ihatdogs10]

sorry

 

[Ihatdogs10]

SIDES PX AND QY
PO YO
OY AND OX

 

[Ihatdogs10]

IS THAT RIGHT?

 

[Harry_the_cat]

Refer to post#90.

IS THAT RIGHT?

They are the sides yes,
Now given that \(\displaystyle \Delta POX\sim\Delta QOY\) write an equivalent statement about the ratio (fractions) of sides. The order of the letters is very important.

 

[Ihatdogs10]

PX/YQ
PO/YO
OX/OQ

Sorry I can't write in fraction

 

[Harry_the_cat]

\(\displaystyle \frac{PO}{QO}=\frac{OX}{OY}=\frac{PX}{QY}\)