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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Abhijeet answered this

Hope it helps.

I'm also giving you a digram:-

Abhijeet answered this

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.

Shrivathsan V answered this

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!

Aditya Kumar Singh answered this

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Madav Mahesh answered this

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Syed Amjad Ali answered this

IF TRIANGLE

Narayani answered this

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