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Illustrative Examples Based on Inequality a Triangle

 

Example 2 :

In ΔABC, AB > AC and D is a point on BC. Show that AB > AD.
 

Solution :

Given: In ΔABC, AB > AC and D is a point on BC.
To Prove: AB > AD.
Proof:
Since in
ΔABC,
AB > AC

... ACB > ABC ...(i)
Now in
ΔADC, ADB is the exterior angle of ΔADC.
ADB = DAC + ACD
ADB > ACD ... (ii)
From inequality (i) and (ii), we get 
ADB > ACD > ABC
 ADB > ABC
 ADB > ABD
 AB > AD ...[side opposite to greater angle]

 

Example 2 :

In the interior of ΔPQR, S is any point, show that PQ + PR > SQ + SR.

Solution :

Given: In the interior of ΔPQR, S is any point.

To Prove: PQ + PR > SQ + SR

Construction: Produce QS to meet PR at T.

 


Proof:
In
ΔPQT, we have
PQ + PT > QT ... [sum of two sides greater than third]

PQ + PT > QS + ST ... (i)
Now in
ΔRST, we have
ST + TR > RS ... (ii)
Adding inequalities (i) and (ii), we get PQ + PT + ST + TR > QS + ST + RS

PQ + (PT + TR) > QS + RS
PQ + PR > QS + RS

 

Example 3 :

In a quadrilateral PQRS, PQ is its longest side and RS is its shortest side. Prove that R > P and  S > Q.

Solution :

Given: In a quadrilateral PQRS, PQ is its longest side and RS is its shortest side.
To Prove :R > P and S > Q.
Construction: Join PR and QS
 


Proof:
In ΔPRS, PS > SR
      3 > 1 ...... (i)
Now in
ΔPQR, PQ > QR
       4 > 2 ...... (ii)
Adding (i) and (ii), we get
           
3 + 4 > 1 + 2
        R > P
Similarly, we can prove that
        
S > Q





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