Points A and B are on the same side of a line m, AD perpendicular to m and BE perpendicular to m and meet m at D and E respectively. If C midpoint of AB then prove that CD=CE.
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Here, AD ⊥ l, CF ⊥ l and BE ⊥ l
AD || CF || BE
In ∆ABE, CG || BE (CF || BE)
And C is the mid-point of AB
Thus, by converse mid-point theorem, G is the mid-point of AE
In ∆ADE, G is the mid-point of AE and GF || AD (CF || AD)
Thus, the converse of mid-point theorem, F is the mid-point of DE.
In ∆CDF and ∆CEF
DF = EF (F is the mid-point of DE)
CF = CF (common)
∠CFD = ∠CFE (Each 90 ° since F ⊥ l)
∴ DDF ≅ DCEF (SAS congruence criterion)
⇒ CD = CE (C.P.C.T)
Hence proved
hopes this helps!!!!.....
AD || CF || BE
In ∆ABE, CG || BE (CF || BE)
And C is the mid-point of AB
Thus, by converse mid-point theorem, G is the mid-point of AE
In ∆ADE, G is the mid-point of AE and GF || AD (CF || AD)
Thus, the converse of mid-point theorem, F is the mid-point of DE.
In ∆CDF and ∆CEF
DF = EF (F is the mid-point of DE)
CF = CF (common)
∠CFD = ∠CFE (Each 90 ° since F ⊥ l)
∴ DDF ≅ DCEF (SAS congruence criterion)
⇒ CD = CE (C.P.C.T)
Hence proved
hopes this helps!!!!.....
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