Question 6
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

Made by

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo

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