Question

Let X be any point on the side BC of ∆ ABC, XM & XN are drawn parallel
to BA & CA. MN meets produced CB in T. Prove that TX2 = TB x TC.

Answer 1

Answer:

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Step-by-step explanation:

Points E and F lie on sides PQ and PR of any △PQR. From the following, for each case find is EF∥QR

(i) PE=3.9cm,EQ=3cm.PF=3.6cm and FR=2.4cm

(ii) PE=4cm,QE=4cm,PF=8cm and RF=9cm

(iii) PQ=1.28cm,PR=2.56cm,PE=0.18cm and PF=0.36cm