□ABCD is a trapezium AB||DC. Points P and Q are midpoints of seg AD and seg BC respectively . Then prove that,PQ||AB and PQ=1/2(AB+DC)
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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)
I HOPE IT WOULD HELP YOU !!!!
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