Question

Prove :
i) AE:AC=AD:AB
ii) DE // BC
iii) triangle PED ~ triangle QCB
pls tell... it's urgent​

Answer 1

Answer:

in AQB AP:AQ=AD:AB

in AQC AP:AQ=AE:AC

hence AE:AC=AD:AB

in ABC AE:AC=AD:AB

then DC//BC

between PED and QCB

PDE=QBC and PED=QCB

henc PED and QCB are similar

since two angle of PED and QCB are equal