Question

In the Figure 5.43,ABCD is a trapezium.AB||DC. Points P and Q are midpoints of segAD and segBC respectively.Then prove that, PQ||AB and PQ =1/2 (AB + DC).

Answer 1

I only know the first half.

Answer 2

Answer:

AB ║ PQ

PQ = (1/2)(AB + DC)

Step-by-step explanation:

Lets join CP & extend it such that it intersects extended BA at point M

so

we get Δ CPD  & Δ MPA

∠DPC = ∠MPA (opposite Angles)

DP = AP  (P is mid point of AD)

∠DCP = ∠AMP  ( as BG ║ DC as G is on extension of BA AB ║ CD)

=> Δ CPD  ≅ Δ MPA

=> MP = CP

& MA = DC

in ΔBCM

MP = CP = CM/2

BQ = CQ = BC/2

now using mid point theorem

AB ║ PQ

& PQ  = (1/2) BM

=> PQ = (1/2) (AM + MA)

=> PQ = (1/2)(AB + DC)