Question

from fig prove that
AE/AC=AD/AB
DE||BC
PED ~ QCB​

Answer 1

Answer

1. In ABQ and ADP, we have,

angle A = angle A. (common)

angle ADP = angle ABQ. (corresponding angles)

ADP ~ ABQ. (by AA criteria )

similarly, APE ~ AQC

SO, AD/AB = AP/AQ

and, AP/AQ = AE/AC

AD/AB = AE/AC. ( i )

2. AD/AB = AE/AC

DE || BC. (by converse of Thales theorem)

as DE || BC and, DP || BQ

hence,

angle PDE = angle QBC. ( corresponding angles)

therefore, proved