Question

In ABCD is a trapezium AB||DC point P andQ are midpoint of seg AD andBC respectively prove PQ|| AB and PQis half (AB-DC)​

Answer 1

Answer:

find it

Step-by-step explanation:

Given,

ABCD is a trapezium in which AB∥DC and P, Q are the midpoints of AD & BC respectively  

Construction : Join CP and produce it to meet AB produced to R.

In ΔPDC & ΔPAR

PD=PA  (∵   P is the midpoint of AD)

∠CPD=∠RPA   (Vertically opposite angles)

∠PCD=∠PRA   (alternate angle)

∴   ΔPDC≅ΔPAR using A as criterion  

⇒CP=CR  (CPCTC)

Also,

In ΔCRB,

P is the midpoint of CR (proved above)

Also, Q is the midpoint of BC

⇒ By midpoint theorem PQ∥AB and PQ=  

2

1

​  

(RB)

But RB=RA+AB

             =CD+AB              (∴   AR=CD by CPCTC)

∴   PQ=  

2

1

​  

(CD+AB)