Perpendicular Bisector

Transcript

Perpendicular BisectorAB is the perpendicular bisector of XY Which means Perpendicular to XY, AB ⊥ XY O is the mid-point of XY, i.e. OX = OY Since we constructed AB Now we try to prove that AB is indeed the perpendicular bisector We do that congruent triangles Proving Perpendicular Bisector To prove AB is perpendicular bisector We prove OX = OY And, ∠ AOX = ∠ AOY = 90° Join A,B to P and A, B to Q In Δ AXB and Δ AYB, AX = AY BX = BY AB = AB ∴ ∆ AXB ≅ ∆AYB Thus, by CPCT (Corresponding Parts of Congruent Triangles) ∠ XAB = ∠ YAB Now , In Δ AXO and Δ AYO, AX = AY ∠ XAO = ∠ YAO AO = AO ∴ ∆ AXO ≅ ∆ AYO Thus, By CPCT (Corresponding Parts of Congruent Triangles) XO = YO Thus, O is mid-point XY Also, by CPCT ∠ AOX = ∠ AOY Now, we need to prove both of them 90° Since XY is a line By Linear Pair ∠ AOX + ∠ AOY = 180° From (2): Putting ∠ AOX = ∠ AOY ∠ AOX + ∠ AOX = 180° 2 × ∠ AOX = 180° ∠ AOX = (180°)/2° ∠ AOX = 90° Thus, ∠ AOX = ∠ AOY = 90° Since XO = YO And, ∠ AOX = ∠ AOY = 90° Thus, we can say AB is the perpendicular bisector of XY