E and F are points on the sides PQ and PR respectively of a ∆P

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180.

E and F are points on the sides PQ and PR respectively of a ∆PQR. For each of the following cases, state whether EF || QR:
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm.
(ii) PE = 4 cm, QE = 4.5 cm. PF = 8 cm and RF = 9 cm.
(iii) PQ = 1.28 cm. PR = 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.


(i) We have,
space PE over EQ equals fraction numerator 3.9 over denominator 3 end fraction equals fraction numerator 1.3 over denominator 1 end fraction space space space space space space space space space space space space space space space space space... left parenthesis straight i right parenthesis
space PF over FR equals fraction numerator 3.6 over denominator 2.4 end fraction equals 3 over 2 equals fraction numerator 1.5 over denominator 1 end fraction space space space space space space... left parenthesis ii right parenthesis
From (i) and (ii), we have
PE over EQ not equal to space PF over FR
Therefore, EF is not parallel to QR [By using converse of Basic proportionality theorem]
(ii) We have,
space PE over QE equals fraction numerator 4 over denominator 4.5 end fraction equals 40 over 45 equals 8 over 9 space space space space... left parenthesis straight i right parenthesis
space PF over RF space equals space 8 over 9 space space space space space space space space space space space space space space space space space space space space space space space space space space space... left parenthesis ii right parenthesis
From (i) and (ii), we have
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Therefore,      EF parallel to QR
 [Using converse of Basic proportionality theorem]

(iii) We have,
PE over PQ equals fraction numerator 0.18 over denominator 1.28 end fraction equals 18 over 128 equals 9 over 64 space space space space... left parenthesis straight i right parenthesis
PF over PR equals fraction numerator 0.36 over denominator 2.56 end fraction equals 36 over 256 equals 9 over 64 space space space space space... left parenthesis ii right parenthesis
From (i) and (ii), we have
                       PE over PQ equals PF over PR
Therefore,       EF parallel to QR
[Using converse of Basic proportionality theorem]
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