# 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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I'm also giving you a digram:- • 50
Here, AD is perpendicular to l, CF is perpendicular to  l and BE is perpendicular to l
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

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.

• 70
The given information can be represented graphically as Here, AD ⊥ l, CF ⊥ l and BE ⊥ l
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

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