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Question
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:
- Point P bisects BE,
- PQ is parallel to AB.
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Solution
The required figure is shown below
(i) From ΔPED and ΔABP,
PD = AP ...[P is the mid-point of AD]
∠DPE = ∠APB ....[Opposite angle]
∠PED = ∠PBA ...[AB || CE]
∴ ΔPED ≅ ΔABP ...[ASA postulate]
∴ EP = BP
(ii) In Δ ECB,
P is a mid point of BE and
Q is a mid point of BC
∴ PQ || CE ...(i) (by mid point theorem)
and CE || AB ... (ii)
From equation (i) and (ii)
PQ || AB
Hence proved.
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