Points * A *and

*are in the same side of a line*

**B***.*

**l***and*

**AD***are perpendiculars to*

**BE***, meeting*

**l***at*

**l***and*

**D***respectively.*

**E***is the*

**C****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!

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