Geometry Problem
Click the figure below to see the complete problem 608.
Saturday, May 28, 2011
Problem 608: Square with Right Triangles, Metric Relations, Congruence, Measurement
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Geometry Problem
Click the figure below to see the complete problem 608.
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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