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Question
In the Figure, `square`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")`.
Solution
Given: `square`ABCD is a trapezium.
To prove: PQ || AB and PQ = `1/2`(AB + DC)
Construction: Extend line AQ in such a way that, on extending side DC, intersect it at point R.
Proof:
seg AB || seg DC ...(Given)
and seg BC is their transversal.
∴ ∠ABC ≅ ∠RCB ...(Alternate angles)
∴ ∠ABQ ≅ ∠RCQ ...(i) ...(B-Q-C)
In ∆ABQ and ∆RCQ,
∠ABQ ≅∠RCQ ...[From (i)]
seg BQ ≅ seg CQ ...(Q is the midpoint of seg BC)
∠BQA ≅ ∠CQR ...(Vertically opposite angles)
∴ ∆ABQ ≅ ∆RCQ ...(ASA test)
seg AB ≅ seg CR ...(c.s.c.t.) ...(ii)
seg AQ ≅ seg RQ ...(c.s.c.t.) ...(iii)
In ∆ADR,
Point P is the midpoint of line AD. ...(Given)
Point Q is the midpoint of line AR. ...[From (iii)]
∴ seg PQ || side DR ...(Midpoint Theorem)
∴ seg PQ || side DC ...(iv) ...(D-C-R)
∴ side AB || side DC ...(v) ...(Given)
∴ seg PQ || side AB ...[From (iv) and (v)]
PQ = `1/2` DR ...(Midpoint Theorem)
= `1/2` (DC + CR)
= `1/2` (DC + AB) ...[From (ii)]
∴ PQ = `1/2` (AB + DC)
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