Answer:
ΔAWM ≅ ΔCPM proved
Explanation:
See the given diagram with this question.
We have to prove ΔAWM ≅ ΔCPM
Now, (i) M is the mid point of AC, so AM =MC
(ii) Now, W is the mid point of AB, hence, AW = WB =

Again, P is the mid point of BC i.e

Now, it is given that BP = PM = MW = WB
Hence, BP = WB , hence, AW = PC
(iii) Again, PM = MW
Therefore, from the SSS rule, it is proved that ΔAWM ≅ ΔCPM. (Answer)