Question

please answer 23rd question

Answer 1

Answer:

given :- Rectangle ABCD a point o inside a rectangle

To prove :- OB^2 + OD^2 = OA^2 + OC^2

proof :-

let us draw PQ through O which is parallel to BC

PQ parallel AD           (opposite side of the rectangle)

all the angles of the rectangle is of 90*

So angle A = angle B = angle C = angle D = 90*

PQ parallel BC & AD is transversal

angle APO = angle B

angle APO = 90*

similarly, we can prove that  

angle BPO = 90*

angle DQO = 90*

angle CQP = 90*

using PT

(hypo)^2 = (height)^2 + (base)^2

In trangle OPB

OB^2 = BP^2 + OP^2                           ...........(1)

In trangle OQD

OD^2 = OQ^2 + DQ^2                              ............(2)

In trangle OQC

OC^2 = OQ^2 + CQ^2                            ..............(3)

In trangle OAP

OA^2 = AP^2 + OP^2                       ................(4)

Adding 1 & 2

OB^2 + OD^2 = BP^2+ OP^2 + OQ^2 + DQ^2

OB^2 + OD^2 =CQ^2 + OP^2+ OQ^2 + AP^2

{As PQ II BC & BP II CQ

BPQC is IIgm

So BP = CQ     (opposite sides of IIgm are equal)

similarly

DQ = AP }

OB^2 + OD^2 =OC^2 + OA^2    ....(from 3&4)

thus   OB^2 + OD^2 =OC^2 + OA^2