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.
Cheers!