Problem 3

In a given right triangle

ABC

, the hypotenuse

BC

has length

a

and is divided into

n

equal parts, where

n

is odd. The central part subtends an angle

\alpha

A

. If

h

is the perpendicular distance from

A

BC

, prove that

\tan\alpha=\dfrac{4nh}{a(n^2-1)}

Step 4 of 5: Invoke Thales' theorem for the midpoint of a right triangle's hypotenuse

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AH^2+MH^2=AM^2=\dfrac{a^2}{4}

Detailed analysis

Since $\angle A=90^\circ$ , the midpoint $M$ of the hypotenuse is the circumcenter, so $AM=BM=CM=a/2$ . In right triangle $AHM$ , this gives $AH^2+MH^2=a^2/4$ .