Geometry Problem 1526. Post your solution in the comment box below.

Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below.

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AG = 20

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ReplyDeleteDefine point I as midpoint of AD and N is the projection of D over AB

Note that AED is the right triangle => IE=IA=ID

Since I is the incenter of triangle AED

So central angle ^(EID)= 2.^(EAD)= 2u=^(NAD)

Triangle IFE similar to AND ( case AA)

So ND/EF=AD/IE= 2

ND=AG=2.EF=20

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ReplyDeleteLet AB, DE extended meet at U. Drop a perpendicular UP to AD, P on AD.

ReplyDeleteTriangles AEU & AED are congruent ASA & so AU = AD

E is midpoint of DU hence from the midpoint theorem UP = 2 X 10 = 20

Now Triangles APU & ADG are congruent ASA (since AU = AD)

So AG = UP = 20

Sumith Peiris

Moratuwa

Sri Lanka

Let K, L be projections of E onto AB, CD, then E being onto the angles bisectors, KE=EF=EL=10, but K-E-L are collinear...

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