108 
Take W as the midpoint of QH and D of PH. Then they are both on the 
nine-point circle and WD must be its diameter, since it is || and equal to 4 QP. 
Then, as 7 Sisaright /, S must also lie on the nine-point circle. S is called 
the vertex of either Simson line. 
13. H’, H”’ and H’”” (Fig. 7) are the points on the circumcircle through 
which AH, BH and CH, respectively, pass. 
U is the point where PH’ cuts BC. 
V is the point where PH” cuts AC, 
W is the point where PH’” cuts AB. 
Now, U, V, H and W lie on a straight line || to the Simson line of P. 
Z VHB» = Z VH” Hy, (Hp is midpoint of HH”.) 
== 7 PE" B = 7 PCs. 
7 UR, UE y= 7 Pn 7 POA: 
Also H, Ha, C, and Hp are concyclic. 
> Z BaHHp 4 C= 180%. 
But we have just proven 2 VHH»+ Z UHHa= Z C. 
‘. Z HaAHHy»+ Z VHA» + UHHa = 180°, and 
.'. U, H and V are collinear. 
Now, 2 WHH’”’ = WH’’H = 180°— 2 PH’’C. 
= Z B+ Z UW’H. 
= / B- 4 DAH": 
Also B, Ha, H, and He are concyclic. 
-. Z B+ Z HaHHe=180° and 
Cee Bee 7 Un 7 rice UI 1808: 
So then 7 WHH’”’” + 4 H-eHU = 180°, which proves W, H and U collinear. 
Therefore all four points, W, V, H and U must be collinear. 
Now, PP, is || to HH”, for both are | to AC. 
. A\ PVX is isosceles, and PP, = P,X. 
Now, Z P,XV= Z PCP, and P, P,, ©, and P,; are concyclic. 
ees ECR == eee 
eo Pr Pek is to Oly Ww. 
From this we can also see that Simson’s line of P bisects all lines from P to 
the line WVHU. 
14. The angle between the Simson lines of two points P and P” is equal to an 
Z inscribed in the circumcircle with PP’ an are and also to an angle inscribed in 
the nine-point circle with arc equal to the part of the circumference included 
between the centers of their Simson’s lines. 
