Question

in rectangle ,ABCD ,if O is an interior point of rectangle ,then prove the following.OA+OB >AB​

Answer 1

To prove:OB

+ OD

=OA

+ OC

Prof. Draw a line PQ || BC

⇒ PQ∣∣AD

All angles of rectangular are 90o

∠A=∠B=∠C=∠D=90

∘

Since, PQ || BC and AB is transversal.

∠APO=∠B (corresponding angles)

∠APO=90

∘

Similarly, ∠BPO=90

∘

, ∠DQO=90

∘

, ∠CQD=90

∘

Using Pythagoras theorem,

In ΔOPB,

(OB)

=(BP)

+(OP)

...(1)

In ΔOQD,

(OD)

=(OQ)

+(DQ)

...(2)

Similarly

(OC)

=OQ

+CQ

...(3)

(OA)

=(AP)

+(OP)

...(4)

Adding (1) and (2)

(OB)

+(OD)

= BP

+OP

+OQ

+DQ

=(CQ)

+(OP)

+(OQ)

+(AP)

=OC

+DA

.

BP=CQ

DQ=AP [Opp. side of ||gram are equal.]

so you satisfied

Answer 2

Step-by-step explanation:

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