# Ex 6.6, 8 (Optional) - Chapter 6 Class 10 Triangles

Last updated at June 15, 2018 by Teachoo

Last updated at June 15, 2018 by Teachoo

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

Chapter 6 Class 10 Triangles

Serial order wise

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.