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Ex 3.4, 6 - ABC is a right-angled triangle and O is the mid point

Ex 3.4, 6 - Chapter 3 Class 8 Understanding Quadrilaterals - Part 2
Ex 3.4, 6 - Chapter 3 Class 8 Understanding Quadrilaterals - Part 3 Ex 3.4, 6 - Chapter 3 Class 8 Understanding Quadrilaterals - Part 4


Transcript

Ex 3.4, 6 ABC is a right-angled triangle and O is the mid point of the side opposite to the right angle. Explain why O is equidistant from A, B and C. (The dotted lines are drawn additionally to help you). Given: A Right-angled triangle ABC. O is the mid-point of AC. To Prove: O is equidistant from A, B and C i.e. OA = OB = OC Construction: Draw a line from A Parallel to BC and Draw a line from C Parallel to BA Let them meet at Point D. Join OD Proof: In ABCD, AB ∥ DC and BC ∥ AD ⇒ Opposite sides are Parallel ∴ ABCD is a parallelogram We know that Adjacent angles of a Parallelogram are Supplementary ∠B + ∠C = 180° 90° + ∠C = 180° ∠C = 180° − 90° ∠C = 90° Also, Opposite angles of a parallelogram are equal. ∠A = ∠C ∠A = 90° Therefore, ∠A = ∠B = ∠C = ∠D = 90° ⇒ Each angle of ABCD is a right angle. So, ABCD is a parallelogram with all angles 90° ∴ ABCD is a rectangle We know that The diagonals of a rectangle bisect each other OA = OC = 1/2 AC OB = OD = 1/2 BD Also, The diagonals of a rectangle are equal in length . BD = AC Dividing both sides by 2 1/2 BD = 1/2 AC OB = OA ∴ OA = OB = OC Hence, O is equidistant from A, B and C. 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.