Check my answers please:
3. For triangle TOE the following facts are given:
mTOS = mSOE
TE = 6 cm
OT = 2 cm
OE = 5.8 cm
OG = 4.35 cm
AU = 0.45 cm
Use this information to answer the following:
a) Why is triangle OBG similar to triangle OTE?
Now find the following missing lengths. Show all work or reasoning. Round non-integral lengths to the nearest hundredth.
e) BT (Use the Side-Splitting Theorem.)
Answers will be in a comment below! Thank you!
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A) First note that both triangles OBG and OTE have the same common angle at vertex O. Therefore, the measure of angle BOG = the measure of angle TOE. Because OS bisects both angles BOG and TOE, it divides TS and SE into the same ratio as OT and OE. Furthermore, because BG is parallel to TE, OS also divides BG = BN + NG into the same ratio. That implies that the sides of the angle BOG are in the same ratio as the sides of angle TOE. Since OB in triangle OBG corresponds to OT in triangle OTE, and OG in triangle OBG corresponds to OE in triangle OTE, then triangles OBG and OTE are similar by equal ratios of corresponding parts.
B) GE = OE - OG
GE = 5.8 cm - 4.35 cm
GE = 1.45 cm
C) TS/SE = OT/OE
TS = SE (OT/OE)
Let TS = x and SE = 6 - x. (Recall that TE = 6.)
x = (6 - x)(2/5.8)
x = (6 - x)(0.344)
x = 6 (0.344) - 0.344x
x = 2.064 - 0.344x
x + 0.344x = 2.064
1.344x = 2.064
x = 2.064/1.344
x = 1.54 to the nearest hundredth
Since TS = x, then TS = 1.54 cm.
D) OA/OU = OG/OE
OA = OU (OG/OE)
OA = (OA + AU)(OG/OE)
OA = (OA + 0.45)(4.35/5.8)
OA = (OA + 0.45)(0.75)
OA = (0.75)OA + 0.3375
OA - (0.75) OA = 0.3375
(0.25)OA = 0.3375
OA = 0.3375/0.25
OA = 1.35 cm
E) BT = OT - OB
BT = 2 - OB
OB/OT = OA/OU
OB = OT(OA/OU)
OB = 2 [1.35/(1.35 + 0.45)]
OB = 2 (0.75)
OB = 1.50
Now plug OB = 1.50 into the equation BT = 2 - OB.
BT = 2 - OB
BT = 2 - 1.50
BT = 0.50 cm
F) SE = TE - TS
SE = 6 - 1.54
SE = 4.46 cm
G) OU = OA + AU
OU = 1.35 + 0.45
OU = 1.80 cm
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that's the basic outline. You can either do a 2 column proof or paragraph proof of that
@jim_thompson5910 Thank you for your help. :)
Everything else looks great. Nice work
@jim_thompson5910 Thank you! I appreciate the help is such a timely fashion. Could I have help with one more proof question? It's essentially the same as question A in the last problem set.
sure, go ahead
For the attached image, would this proof (based off the proof you gave me earlier) be correct?
AN || RI (Given.)
angle ATN = angle RTI (Common angle.)
angle TAN = angle TRI (Corresponding angle.)
triangle TAN ~ triangle TRI (AA Similarity theorem.)