Check sibling questions

Ex 6.6, 8 (Optional) - Prove that (i) PAC ~ PDB (ii) PA . PB = PC . PD

Ex 6.6, 8 (Optional) - Chapter 6 Class 10 Triangles - Part 2
Ex 6.6, 8 (Optional) - Chapter 6 Class 10 Triangles - Part 3 Ex 6.6, 8 (Optional) - Chapter 6 Class 10 Triangles - Part 4

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Transcript

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 PAC PDB Given: A Circle with two chords AB and CD which meet at a Point P outside the circle. To Prove: (i) PAC 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. PB = PC.PD 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.