Geometry Problem

Click the figure below to see the complete problem 608.

## Saturday, May 28, 2011

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## Saturday, May 28, 2011

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Problem 608: Square with Right Triangles, Metric Relations, Congruence, Measurement

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congruence,
measurement,
metric relations,
right triangle,
square

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Let EJ⊥AD and FK⊥CD

ReplyDeleteΔEAD ≡ ΔFDC

∴ EJ = FK

Now HM/HN = HE/HC = EJ/CD = FK/AD = GF/GA = GN/GM

Hence 1 – MN/HN = 1 – MN/GM

Follows HN = GM,

HM = GN

excellence solution

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