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Mathematics examination paper for senior high school entrance examination in Xining in 2005
Solution: extend BD to m so that MA is perpendicular to BA.

It is easy to prove that triangle CEA is equal to triangle BMA, so EA=AM.

Triangle BHE is similar to triangle BMA.

So BE:BA=HE:MA BA*HE=BE*MA.

Let: OE=x, so EA = AM =1-xbe =1+X.

2*HE=( 1+x)( 1-x)

He =( 1-x squared) /2

According to Pythagorean theorem, we can get OH=( 1+x squared) /2.

So oh+he = 1.