Question

E and F are points on the sides PQ and PR respectively of a ΔPQR. Show that EF || QR, if PQ = 1.28 cm, PR= 2.56 cm, PE = 0.18 cm and PF = 0.36 cm.

Answer 1

We have, PQ = 1.28 cm, PR = 2.56 cm
PE = 0.18 cm, PF = 0.36 cm
Now, EQ = PQ – PE = 1.28 – 0.18 = 1.10 cm
and FR = PR – PF = 2.56 – 0.36 = 2.20 cm
Now, PE/EQ= 0.18/1.10 =18/110 = 9/55

and,
PF/FR = 0.36/2.20 = 36/220 = 9/55 = PE/EQ = PF/FR
Therefore, EF || QR [By the converse of Basic Proportionality Theorem]