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Which information could he have used to determine this?

O GLH = ILM
O m KLM = 5m ILMO
O m GLI = 2m GLH
O m GLI = 1/2m GLH + 1/2m HLI​


Which Information Could He Have Used To Determine ThisO GLH ILMO M KLM 5m ILMOO M GLI 2m GLHO M GLI 12m GLH 12m HLI class=

Sagot :

Answer:

m∠GLI = 2m∠GLH

Step-by-step explanation:

Angle GLI (red) is made of angles GLH (purple) and HLI (green).

m∠GLI = m∠GLH + m∠HLI

First option: ∠GLH ≅ ∠ILM (Angle GLH is congruent to angle ILM.)

This doesn't tell us anything about the relationship between angles GLI, GLH and HLI. So it can't be used.

Second option: m∠KLM = 5m∠ILM (Measure of angle KLM is 5 times the measure of angle ILM.)

This also cannot be used, because it doesn't include angles GLI, GLH or HLI.

Third option: m∠GLI = 2m∠GLH (Measure of angle GLI is two times the measure of angle GLH.)

Now this option is more interesting. It includes two angles of three that we are interested in.

If angle GLI is two times angle GLH, then that means that angle GLH is congruent to angle HLI!

Let's prove this with equations.

1) m∠GLI = m∠GLH + m∠HLI (from picture)

2) m∠GLI = 2m∠GLH (from third option)

Substitute m∠GLI in first equation with m∠GLI in second.

m∠GLI = m∠GLH + m∠HLI

2m∠GLH = m∠GLH + m∠HLI

m∠GLH = m∠HLI

So this proves that the angles GLH and HLI have the same measure, therefore ray LH is bisector of GLI.

Fourth option: m∠GLI = 1/2m∠GLH + 1/2m∠HLI​

There are all angles we are interested in, but there is a mistake. We can see in the picture that angle GLI is made of angle GLH and HLI and not of half of them. This just is not true. The correct equation would be m∠GLI = m∠GLH + m∠HLI.

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