Points A and B are in the same side of a line l. AD and BD are perpendiculars to l, meeting at D and E. C is the midpoint of AB. Prove that CD = CE.

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Here, AD is perpendicular to l, CF is perpendicular to  l and BE is perpendicular to l
AD || CF || BE
In Triangle 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 Triangle 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 Triangles CDF and CEF
DF = EF  (F is the mid-point of DE)
CF = CF  (common)
Angle CFD = Angle CFE   (Each 90 degrees since F is perpendicular to l)
∴ DDF is congruent to DCEF  (SAS congruence criterion)
=> CD = CE  (C.P.C.T)
Hence proved.
 
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The given information can be represented graphically as
 
 
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
 
Hope! This will help you.
Cheers!
 
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