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Example 14 - O is any point inside a rectangle ABCD - Examples

Example 14 - Chapter 6 Class 10 Triangles - Part 2
Example 14 - Chapter 6 Class 10 Triangles - Part 3
Example 14 - Chapter 6 Class 10 Triangles - Part 4

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Example 14 O is any point inside a rectangle ABCD (see Fig. 6.52). Prove that OB2 + OD2 = OA2 + OC2. Given : Rectangle ABCD , and a point O inside rectangle . To prove :- OB2 + OD2 = OA2 + OC2 Proof :- Let us draw a line PQ, through O which is parallel to BC. Hence, PQ II BC ⇒ PQ II AD All angles of a rectangle are 90° , So, ∠ A = ∠ B = ∠ C = ∠ D = 90° Since, PQ II BC & AB is the transversal ∠ APO = ∠ B ∠ APO = 90° Similarly, we can prove ∠BPO = 90° , ∠ DQO = 90° & ∠CQO = 90° Using Pythagoras theorem. (Hypotenuse)2 = (Height)2 + (Base)2 Similarly, In right triangle ∆ 𝑂𝑄𝐶 , OC2 = OQ2 + CQ2 …(3) & In right triangle ∆ 𝑂𝐴𝑃 , OA2 = AP2 + OP2 …(4) Adding equation (1) and (2) OB2 + OD2 = BP2 + OP2 + OQ2 + DQ2 = (CQ)2 + (OP)2 + (OQ)2 + (AP)2 = CQ2 + OQ2 + OP2 + AP2 = OC2 + OA2 Thus, OB2 + OD2 = OC2 + OA2 Hence proved

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Davneet Singh

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.