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  1. Chapter 6 Class 10 Triangles
  2. Serial order wise
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Ex 6.6, 8 In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that Δ PAC ∼ Δ PDB Given: A Circle with two chords AB and CD which meet at a Point P outside the circle. To Prove: (i) ∆ PAC ∼ Δ PDB Proof: We know that, ABDC is a cyclic quadrilateral, ∴ Sum of its opposite angle is 180° Thus, ∠BAC + ∠PDB = 180° Also, ∠BAC + ∠PAC = 180° ∠BAC = 180° − ∠PAC Putting ∠BAC = 180 − ∠PAC in equation (1) ∠BAC + ∠PDB = 180° 180 − ∠PAC + ∠PDB = 180° ∠PDB = 180° − 180 + ∠PAC ∠PDB = ∠PAC In ∆PAC and ∆PDB, ∠APC = ∠BPD ∠PAC = ∠PDB ∴ ∆ PAC ∼ ∆ PDB Hence Proved. Ex 6.6, 8 In Fig. 6.62, two chords AB and CD of a circle intersect each other at the point P (when produced) outside the circle. Prove that (ii) PA . PB = PC . PD From part (i) ∆ PAC ∼ ∆ PDB Thus, 𝑃𝐴/𝑃𝐷 = 𝑃𝐶/𝑃𝐵 PA. PB = PC.PD Hence Proved

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