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Example 5 - Show that bisectors of angles of parallelogram - Opposite angles of parallelogram

Example 5 - Chapter 8 Class 9 Quadrilaterals - Part 2
Example 5 - Chapter 8 Class 9 Quadrilaterals - Part 3

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Example 5 Show that the bisectors of angles of a parallelogram form a rectangle. Given: ABCD is a parallelogram AP, BP, CR, DR are bisectors of ∠ A , ∠ B, ∠ C, ∠ D respectively To prove: PQRS is a rectangle Proof: A rectangle is a parallelogram with one angle 90° First we will prove PQRS is a parallelogram Now, AB ∥ DC & AD is transversal ∴ ∠ A + ∠ D = 180° Multiplying by half 1/2∠ A + 1/2∠ D = 1/2 × 180° 1/2∠ A + 1/2∠ D = 90° ∠ DAS + ∠ ADS = 90° Now, In Δ ADS ∠ DAS + ∠ ADS + ∠ DSA = 180° 90° + ∠ DSA = 180° ∠ DSA = 180° – 90° ∠ DSA = 90° Also, lines AP & DR intersect So,∠ PSR = ∠ DSA ∴ ∠ PSR = 90° Similarly, we can prove that ∠ SPQ = 90° , ∠ PQR = 90° and ∠ SRQ = 90°. So, ∠ PSR = ∠ PQR & ∠ SPQ = ∠ SRQ ∴ Both pair of opposite angles of PQRS are equal So, PQRS is a parallelogram Also, ∠ PSR = ∠ PQR = ∠ SPQ = ∠ SRQ = 90°. ∴ PQRS is a parallelogram in which one angle 90° ⇒ PQRS is a rectangle. 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.