Q22 of 47 Page 7

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

Given: ΔABC in which AD is a median.


To prove: AB+AC>2AD


Construction: Produce AD to E such that AD = DE. Join EC.



Proof:


In Δ ADB and ΔEDC,


AD = DE (By construction)


BD = DC (∵ D is midpoint of BC)


∠ADB = ∠EDC (Ver. Opp. Angles)


By SAS congruency,


Δ ADB ≅ ΔEDC


AB = EC (By CPCT)


In ΔAEC,


As sum of two sides is greater than third side.


AC + EC > AE


∵ AD = DE


∴ AE = AD + DE and EC = AB


⇒ AC + AB > 2 AD


Hence proved


More from this chapter

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20

ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE=CF, prove that ΔABC is isosceles.

 21

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

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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)