On the hypotenuse $AB$ of triangle $ABC$, with $\angle C =90^{\circ}$ and area $S$, as on the diameter, was drawn a circle. The points $K$ and $M$ were chosen on arcs $AB$ and $AC$ respectively, in such a way that chord $KM$ is a diameter of the circle. Let $P$ and $Q$ be the bases of the perpendiculars, drawn from points $A$ and $C$ on lines $CM$ and $AM$ respectively. Prove, that the area of $KPMQ$ equals $S$.
I. Nagel
Construction: 1) PQ is joined and it cuts KOM extended at X. 2) CD is perpendicular to AB at D. 3)PD is joined and it cuts AMQ at Y.
Claim#1: PXMY is a cyclic quadrilateral.
Proof: $\delta = \alpha + \beta = \alpha ‘ + \beta ‘$.
Claim#2: $PX$ is the altitude of $\triangle PMK$ using $MK$ as base.
Proof: $\alpha ‘ + \theta = \alpha + \theta_1 = \alpha + \theta_2 = 90^\circ$
This means $\angle PXM = 90^0 \circ$
Direct consequences from the above are:-
1] $QX$ is the altitude of $\triangle QMK using $MK$ as base.
2] $QC // PD$ because $\angle QYD = \angle PXM = 90^0 \circ$ and $\angle CQM$ is also $90^\circ$
3] PQ = CD because QC and PD are parallel chords of the same circle.
$[KPMQ] = [\triangle PMK] + [\triangle QMK] = … = \dfrac {1}{2} PQ \cdot MK = \dfrac {1}{2} CD \cdot MK = \dfrac {1}{2} CD \cdot AB = S$