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If two angles are equal and have a pair of sides in coincidence, then the 
other sides must also either coincide or be parallel. Hence AU || P, P, Ps, or to 
Simson’s line. Thus we can show BU, and CU, parallel to Simson’s line of P 
and therefore AU,, BU, and CU, are parallel to each other. 
3. Let AT (Fig. 1) be isogonal conjugate to AP. Then Simson’s line of 
Pop AT. 
Also Simson’s line of T | AP. 
Now, AU, is || Simson’s line of P, and 
Zot 7 PAC 180° — 4. PU,C. 
Aiso.7- BAU, = Z BCU. 
-, Z BAT— Z PAU, =180°— 2 PU,C— Z BCU,. 
.". U,AT = 180° — 90° = 90°. for Z PU,C is measured by 3 arc PC and 
Z BCU, is measured by 3 are BU,. 
But PP,C, which is a right /, is measured by 3 arc ( PC + BU,). 
... U,A | AT and so Simson’s line of P must be. In like manner we can 
prove Simson’s line of T | AP. 
Now, if Q is the point on the cireumference opposite P, then AU, and AQ 
are isogonal conjugate lines, for 
Za A == / QAP = 90° and 
Z TAC= Z BAP with Z U,AQ common. 
, Z U,AT— Z QAU, — Z TAC= Z QAP— Z U,AQ— Z BAP. 
eA == 7 CAQ. 
4. If P and Q are opposite points on the circumference, their Simson’s lines 
are | to each other. 
Now, the isogonal conjugate of AP is | to isogonal conjugate of AQ, and, 
therefore, since the Simson’s line of P | AU, and the Simson’s line of Q | AT, the 
Simson’s line of P will be | to Simson’s line of Q. 
5. A side, BC, and its altitude in a triangle are the Simson’s lines of A” 
and A, respectively, where A’ is the point on the circumference opposite A. 
( Fig 4.) 
Since the feet of the | s from A to AB and AC coincide with A, and the foot 
of the | from A to BC is Ha, therefore the Simson’s line of A is AHg. Again, the 
feet of |. s from A’ to the sides AB, AC, and BC are B, C, and A/a, respectively ; 
hence BC is the Simson’s line of A’. 
Since AHa, BH» and CH¢ are the Simson’s lines of A, B, and C, respectively, 
their Siftson’s lines concur in H, the ortho-center. 
6. Let A’, B’ and C’ be the points on the circumference opposite A, B and O, 
respectively, and Ha’, Hy’, and H-’ be the points where AHa, BHa and CHa, 
produeed, cut the circumference, then the 
