Question

12. In trapezium ABCD, AB is parallel to DC;
P and Q are the mid-points of AD and BC
respectively. BP produced meets CD
produced at point E.
Prove that :
(i) Point P bisects BE,
(ii) PQ is parallel to AB.​

Answer 1

Answer:

Step-by-step explanation:

Proved that point P bisects BE  & PQ ║ AB  in Trapezium AB ║ DC  , P & Q mid points of AD & BC,   BP produced meets CdD Produced at E

Step-by-step explanation:

in ΔPED  & ΔPBA

PD = PA   ( as P is mid point of DA)

∠EPD = ∠BPA  ( opposite angle)

∠PDE = ∠PAB  ( as CD ║ AB)  

ΔPED  ≅ ΔPBA

PE = BP

=>  point P bisects BE

in Δ EBC & ΔPBQ

EB/PB = BC/BQ = 2

=> PQ ║ CE

=> PQ ║ CD

=>  PQ ║ AB

draw the figure by your own it is only

Answer 2

Answer:

here is your answer

Step-by-step explanation:

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