Question

R Prove that: (1)
trapezium ABCD, AB is parallel to DC; P
and Q are the mid-points of AD and BC
spectively. BP produced meets CD produced
at point E. Prove that :
(1) point P bisects BE
(i) PQ is parallel to AB.​

Answer 1

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

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Answer 2

Answer with Step-by-step explanation:

AB is parallel to DC

P is the midpoint of AD and Q is the mid- point of BC.

1.In triangle EDP and triangle APB

AP=PD (P is the midpoint of AD)

Angle EPD=Angle APB

Reason: vertical opposite angles are equal

Angle ADP=Angle ABP

Reason: Alternate interior angle

$\triangle$EPD$\cong\triangle APB$

Reason: ASA postulate

AP=PB

Reason: CPCT

Therefore, P is the mid-point of BE.

2.In triangle BEC

$\frac{EP}{PB}=1$

$\frac{CQ}{QB}=1$

$\frac{EP}{PB}=\frac{CQ}{QB}$

$PQ\parallel CE$

Reason: By converse of proportionality theorem

AB is parallel to CD ( CD is a part of CE)

AB is parallel to CE

Therefore, PQ is parallel to AB.

Hence, proved.

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