Q25 of 47 Page 7

In Fig. , prove that:


(i) CD + DA + AB + BC > 2AC


(ii) CD+DA +AB > BC

Proof:



From the given figure,


We know that,


In a triangle sum of any two sides is greater than the third side.


(i) So,


In ∆ABC, we have


AB + BC > AC (1)


In ∆ADC, we have


CD + DA > AC (2)


Adding (1) and (2), we get


AB + BC + CD + DA > AC + AC


CD + DA + AB + BC > 2 AC


(ii) Now, in ∆ABC, we have


CD + DA > AC


Add AB on both sides, we get


CD + DA +AB > AC + AB > BC


CD + DA + AB > BC


Hence, proved


More from this chapter

All 47 →
21

In Fig., AD ⊥ CD and CB ⊥ CD. If AQ=BP and DP =CQ, prove that ∠DAQ = ∠CBP.

 22

Prove that any two sides of a triangle are together greater than twice the median drawn to the third side.

 23

If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.

 24

O is any point in the interior of Δ ABC. Prove that

(i) AB + AC > OB + OC


(ii) AB + BC + CA > OA + OB + OC


(iii) OA + OB + OC > (AB + BC + CA)