Question

E and F are respectively the midpoints of the the non parallel sides AD and BC of a trapezium ABCD prove that

(i) EF || AB,
(ii) EF = ½ + (AB + CD).​

Answer 1

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$\huge\underline\bold\orange{Question}$

E and F are respectively the midpoints of the the non parallel sides AD and BC of a trapezium ABCD prove that

(i) EF || AB,

(ii) EF = ½ + (AB + CD).

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$\huge\underline\bold\orange{Solution}$

${\blue{\sf\underline{Given}}}$

A trapezium ABCD in which E and F are midpoints of sides AB and BC respectively.

${\blue{\sf\underline{To\:prove}}}$

$(i) \: ef || ab$

$(ii) \: ef = \frac{1}{2} (ab + cd).$

${\blue{\sf\underline{Construction}}}$

Join DF and produce it to meet AB produced in P.

${\blue{\sf\underline{Proof}}}$

In ∆DCF and PBF we have

⠀⠀⠀⠀⠀⠀∠DFC = ∠PFB (vert.opposite ∠s)

⠀⠀⠀⠀⠀⠀CF = BF (∴ F is the midpoint of BC)

⠀⠀⠀⠀⠀⠀∠DCF = ∠PBF (alt interior ∠s)

∴ ∆DCF ≅ ∆PBF (ASA-criterioj).

And so, DF = PF and CD = BP (c.p.c.t.).

Now,in ∆DAP,we have

E is the midpoint of AD and F is the midpoint of DP. [∴ DF = PF]

$\therefore \: EF || AP \: and \: Ef = \frac{1}{2} \: (by \: midpoint \: theorem)$

$⇒ EF || AB \: and \: EF = \frac{1}{2} (AB + BP)$

$hence \: EF || AB \: and \: EF = \frac{1}{2} (AB + CD)$

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Answer 2

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$<$

Question - E and F are respectively the midpoints of the the non parallel sides AD and BC of a trapezium ABCD prove that

(i) EF || AB,

(ii) EF = ½ + (AB + CD).

❤▬▬▬▬Solution▬▬▬▬❤

Given - A trapezium ABCD in which E and F are midpoints of sides AB and BC respectively.

$(i)ef∣∣ab</p><p></p><p>(ii) \: ef = \frac{1}{2} (ab + cd).(ii)ef=21(ab+cd).</p><p></p><p>{\blue{\sf\underline</p><p></p><p></p><p>$

Construction➽ - Join DF and produce it to meet AB produced in P.

Proff - In ∆DCF and PBF we have

⠀⠀⠀⠀⠀⠀∠DFC = ∠PFB (vert.opposite ∠s)

⠀⠀⠀⠀⠀⠀CF = BF (∴ F is the midpoint of BC)

⠀⠀⠀⠀⠀⠀∠DCF = ∠PBF (alt interior ∠s)

∴ ∆DCF ≅ ∆PBF (ASA-criterioj).

And so, DF = PF and CD = BP (c.p.c.t.).

➥➥➥Now,in ∆DAP,we have➥➥➥

E is the midpoint of AD and F is the midpoint of DP. [∴ DF = PF]

$EF∣∣APandEf=21(bymidpointtheorem)</p><p></p><p>\begin{gathered}⇒ EF || AB \: and \: EF = \frac{1}{2} (AB + BP) \\\end{gathered}⇒EF∣∣ABandEF=21(AB+BP)</p><p></p><p>\begin{gathered}hence \: EF || AB \: and \: EF = \frac{1}{2} (AB + CD) \\\end{gathered}henceEF∣∣ABandEF=21(AB+CD)</p><p></p><p></p><p>$

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