plz help me with the solution of question no.23
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Step-by-step explanation:In ∆ EAD & ∆DCF,
∠1 = ∠ 2 [ corresponding angles are equal , as AB || DC]
∠3 = ∠ 4 [ corresponding angles are equal , as AD || BC]
∆ EAD ~ ∆DCF
[By AA similarity criterion]
EA/DC = AD/CF = DE/FD
[corresponding sides of similar triangles are proportional]
AD/AE = CF/CD…………….(1)
In ∆ EAD & ∆EBF,
∠1 = ∠ 1 (Common angle)
∠3 = ∠ 4 [ corresponding angles are equal , as AD || BC]
∆ EAD ~ ∆EBF
[By AA similarity criterion]
EA/EB = AD/BF = DE/EF
AD/AE= FB/BE……………….(2)
From eq 1 & 2
AD/AE= FB/BE = CF/CD
Hence, proved.
HOPE THIS WILL HELP YOU..
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