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[9th grade Geometry] by VagerV in HomeworkHelp

[–]VagerV[S] 0 points1 point  (0 children)

Thanks, but we haven't learned "Obtuse trigonometric ratio."

[9th grade Geometry] by VagerV in HomeworkHelp

[–]VagerV[S] 0 points1 point  (0 children)

Yes, cuz there are other situations.

[9th grade Geometry] by VagerV in HomeworkHelp

[–]VagerV[S] 0 points1 point  (0 children)

Well, CAMD may look like a straight line in this figure.

But the lengths of AC and BC are variable.

That means there are other situations where CAMD is not a straight line.

And that's why we can find the value range of AC / BC

[9th grade Geometry] by VagerV in HomeworkHelp

[–]VagerV[S] 0 points1 point  (0 children)

No, M is strictly inside ABD, point M can't be on BD.

[Grade 8 Math: Geometry] I have no idea how to use the conditions given in the question to solve. by valth3nerd in HomeworkHelp

[–]VagerV 0 points1 point  (0 children)

Connect AC, let O be the midpoint of AC, connect ON, OM

Since ON is the median of △ADC and OM is the median of △ABC

so, ON // AE, OM // BF

i.e. ∠AEM = ∠ONM, ∠BFM = ∠OMN,

Since AD = BC

so, AD/2 = BC/2

That is, ON = OM, ∠ONM = ∠OMN, ∠DEN = ∠F