The correct option is
A True
In
△s, APB and ECB,
∠ABP=∠EBC (Common angle)
∠PAB=∠CEB (Corresponding angles of parallel lines)
∠APB=∠ECB (Third angle of the triangle)
Thus △APB∼△ECB (AAA rule)
Hence, ABEB=BPBC (Corresponding sides of similar triangles)
2=BPBC
BP=2BC
Now, in △s OPC and APB,
∠OPC=∠APB (Common angle)
∠POC=∠PAB (Corresponding angles of parallel lines)
∠PCO=∠PBA (Third angle of a triangle)
△OPC∼△APB (AAA rule)
hence, PCBP=OPAP (Corresponding sides)
12=OPAP
OP=12AP
hence, O is the midpoint of AP.
![198211_179107_ans_e2a4a110d7cd4980979a0949cf5580a3.png](https://search-static.byjusweb.com/question-images/toppr_ext/questions/198211_179107_ans_e2a4a110d7cd4980979a0949cf5580a3.png)