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Example 7 - Show that the line PQ is the perpendicular - Examples

Example 7 - Chapter 7 Class 9 Triangles - Part 2
Example 7 - Chapter 7 Class 9 Triangles - Part 3

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Example 7 AB is a line-segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (see figure). Show that the line PQ is the perpendicular bisector of AB. Given: P is equidistant from points A & B PA = PB and Q is equidistant from points A & B QA = QB To prove: PQ is perpendicular bisector of AB, i.e. AC = BC & ∠ PCA = ∠ PCB = 90° Proof : In ∆ PAQ and ∆ PBQ. AP = BP AQ = BQ PQ = PQ So, Δ PAQ ≅ Δ PBQ Hence, ∠ APQ = ∠ BPQ In Δ PAC and Δ PBC. AP = BP ∠ APC = ∠ BPC PC = PC Δ PAC ≅ Δ PBC Therefore, AC = BC and ∠ ACP = ∠ BCP ∠ ACP = ∠ BCP …(4) Since, AB is a line segment, ∠ ACP + ∠ BCP = 180° ∠ ACP + ∠ ACP = 180° 2∠ ACP = 180° ∠ ACP = 180"°" /2 ∠ ACP = 90° So, ∠ ACP = ∠ BCP = 90° Thus, AC = BC & ∠ ACP = ∠ BCP = 90° ∴ PQ is perpendicular bisector of AB. Hence proved

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.