SAT Math 3.3 Geometry and Measurement
|Additional Topics in Math||Congruence and similarity|
|Geometry and Measurement||Points and Lines|
|Product Type||SAT Math|
|SAT Math||Geometry and Measurement|
And here are the potential answers:
Okay, yeah – we get the warning that the figure was “not drawn to scale”
…but it's like this diagram was apparently drawn by a six year old.
We’re going to draw our own diagram. And it’s going to be amazing.
BD bisects our original angle, so ABD and DBC are equal.
BF bisects DBC, so both DBF and FBC must be one-half the measure of n. So far so good.
Same deal with BG, which bisects FBC… which means that now both FBG and GBC are one-fourth
the measure of n.
Now here's where it gets tricky. BE bisects DBG.
Well, we know the measure of DBC – one n… and we know the measure of GBC – one-fourth n.
The angle DBG must then be the difference – or three-fourths n.
Because BE is bisecting THAT angle… EBC must be half of three-fourths – or three-eighths
– plus angle GBC right here.
Three-eighths plus one-fourth – or, simplified, three-eighths plus two-eighths, is five-eighths.
And that's it...Answer D.