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Example 6 - Prove that quadrilateral formed by internal - Examples

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  1. Chapter 10 Class 9 Circles
  2. Serial order wise
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Example 6 Prove that the quadrilateral formed (if possible) by the internal angle bisectors of any quadrilateral is cyclic. Given: ABCD is a quadrilateral AH, BF, CF, DH are bisectors of ∠ A , ∠ B, ∠ C, ∠ D respectively To prove: EFGH is cyclic quadrilateral Proof: To prove EFGH is a cyclic quadrilateral, we prove that sum of one pair of opposite angles is 180° In Δ AEB ∠ ABE + ∠ BAE + ∠ AEB = 180° ∠ AEB = 180° – ∠ ABE – ∠ BAE ∠ AEB = 180° – (1/2 ∠ B + 1/2 ∠ A) ∠ AEB = 180° – 1/2 (∠ B + ∠ A) Now, lines AH & BF intersect So, ∠ FEH = ∠ AEB ∴ ∠ FEH = 180° – 1/2 (∠ B + ∠ A) Similarly, we can prove that ∠ FGH = 180° – 1/2 (∠ C + ∠ D) Adding (2) & (3) ∠ FEH + ∠ FGH = 180° – 1/2 (∠ A + ∠ D) + 180° – 1/2 (∠ C + ∠ B) ∠ FEH + ∠ FGH = 180° + 180° – 1/2 (∠ A + ∠ D + ∠ C + ∠ B ) ∠ FEH + ∠ FGH = 360° – 1/2 (∠ A + ∠ B + ∠ C + ∠ D ) ∠ FEH + ∠ FGH = 360° – 1/2 (∠ A + ∠ B + ∠ C + ∠ D ) ∠ FEH + ∠ FGH = 360° – 1/2 × 360° ∠ FEH + ∠ FGH = 360° – 180° ∠ FEH + ∠ FGH = 180° Thus, in EFGH, Since sum of one pair of opposite angles is 180° EFGH is a cyclic quadrilateral

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
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