Points A and B are in the same side of a line l . AD and BE are perpendiculars to l , meeting l at D and E respectively. C is the mid point of line segment AB .

Proove that  CD = CE .

Here is the answer to your query.

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.


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