15.

Given: sPC @ sQC
            A is midpt. of sPQ
            ÐPCB @ ÐQCB
Prove: line BA ^ line PQ
[More interesting: prove that line PQ ^ m.]

1. sPC @ sQC

1. Given

2. ÐPCB @ ÐQCB

2. Given

3. sCB @ sCB

3. Refl.

4. DPCB @ DQCB

4. SAS (steps 1, 2, 3)

5. sPB @ sQB

5. CPCTC

6. B is on a ^ bisector of sPQ

6. Equidistance Thm. (a.k.a. ^ Bis. Thm.), from step 5

7. C is on a ^ bisector of sPQ

7. Equidistance Thm. (a.k.a. ^ Bis. Thm.), from step 1

8. A is midpt. of sPQ

8. Given

9. sAB is a ^ bisector of sPQ;
    sAC is a ^ bisector of sPQ

9. Def. ^ bisector, plus steps 6-8

10. sAB ^ line PQ;
      sAC ^ line PQ

10. Def. ^ bisector

You could stop here, but why not continue and show that line PQ is ^ to the entire plane?

11. ÐCAB is not a straight Ð

11. Def. of D (DABC shown in diagram)

12. line AB and line AC are distinct lines lying in plane m and intersecting at foot A

12. Step 11, diagram

13. line PQ ^ m

13. Thm.: A line ^ to 2 distinct lines in a plane that pass through its foot is ^ to the plane.