4.4 Perpendicular Lines Date: 1) Construct Perpendicular Bisector 2) Construct a perpendicular line through a point not on the line 3) Recall: Construct a line parallel to the given line through the given point- Use any method! 1) Proving the Perpendicular Bisector Theorem using Transformations 2) Proving the Converse of the Perpendicular Bisector Theorem using Indirect Reasoning (or Proof by Contradiction) Given: is on the perpendicular bisector of Prove: PA= PB Consider the reflection across _________. Then the reflection of point P across line m is also ___________ because point P lies on _________, which is the line of reflection. Also, the reflection of _______ across line m is B by the definition of __________. Therefore PA = PB because __________ preserves distance. Reflect: What can you conclude about ∆? Given: PA = PB Prove: is on the perpendicular bisector of Step 1: Assume what you are Trying to prove is false. Assume Then when you draw a perpendicular line from P containing A and B, it intersects at point Q, which is not ___________________. PQ forms two right triangles. __________ and __________ So by Pythagorean theorem, Contradiction because Z A B C B And key Job . Ntt 10/20/16 " %t* to Tammy .mn#*BYYedisian*tnabdiYrtYutms*endpts.(Maethanhalfway)tBiseokrTheorem : Converse : lfaptiseqvidistantfrmendpts . Ifapointisanthebbiseotorofasegment , qasegmentthenitismtbisector . then it is equidistant famendpts . of indirect : Assumewnatyaiwanttopweis of the segment false agettoa contradiction ÷ inem pisnotmthesbiseutrm AATB p linen A themidptoftb reflection reflection OPQA OPQB PQ2tAQ2=pA2 And POITQBYPBZ PQrtAQ2=PQ2tQB2 okljisisosaleb NANNY AQ=QBf HMJ⇒lMJ because okmjmappedtoowp We said QWASNOIMIDPHOFAB . by a reflection . Somyopotc .ws#wwhsqPmustbemtbseCtNMofABmaHsitisosaksby definition