Let AXYZB be a convex pentagon inscribed in a semicircle of diameter AB. Denote by P, Q, R, S the feet of the perpendiculars from Y onto lines AX, BX, AZ, BZ, respectively. Prove that the acute angle formed by lines PQ and RS is half the size of ∠XOZ, where O is the midpoint of segment AB.

Call the intersection of PQ and SR point K
Call the intersection of AZ and BX point L
Call the measure of ∠AOX = arc angle AX = x
Call the measure of ∠BOZ = arc angle BZ = z

∠AXB = ∠BZA = 90 since they are inscribed in a semicircle.   Given the perpendiculars in the problem, quadrilaterals XPYQ and ZSYR are rectangles.

It is simple to establish the following:
∠XOZ = arc XZ = 180-x-z
∠ZAB = z/2
∠XBA = x/2
∠ALB = 180 - x/2 - z/2 = ∠XLZ
∠QLA = ∠RLB = x/2 + z/2
∠QYR = x/2 + z/2

Since the arc XZ = 180 - x - z
the two inscribed angles ∠YXZ and ∠YZX sum to half of this: 90 - x/2 - z/2
and so ∠XYZ = 90 + x/2 + z/2
subtracting ∠QYR tells us that ∠XYQ and ∠ZYR are complimentary.
Thus ∠PQY and ∠SRY are complimentary.
And since ∠YQL and ∠YRL are right, ∠KQL and ∠KRL are also complimentary.

Finally consider the concave quadrilateral KQLR.
Two of its angles sum to 90 and the reflex angle QLR = (x/2 + z/2) + (180 - x/2 - z/2) + (x/2 + z/2) = 180 + x/2 + z/2 leaving for the fourth ∠PKS = 90 - x/2 - z/2 which is one half of ∠XOZ as required.  QED

Incidentally:
The rectangles are similar, which is easy to see.
Point K is on AB (It appears to be at least.  I didn't prove it.)

Posted by Jer on 2014-01-12 01:33:18