Question

We have a triangle ABC. Point D lies on BC and point E lies on the side AB. We know that, AC = BD , AD = AE , AB2 = (AC)(BC). How can you prove that angle BAD = angle CEA ?

Answer 1

Given

`"For "Delta ABC " we have"`

`AC=BD," " AD=AE" and "AB^2=AC*BC`

`"To prove " /_BAD=/_CEA`

Proof

`"From given condition "AB^2=AC*BC`

`=>AB^2=AC*BC=BD*BC," since "AC=BD`

`=>(AB)/(BC)=(BD)/(AB).............................(1)`

The relation (1) reveals that

`DeltaABC and Delta ABD" are similar"`

`"Their corresponding angles are"`

`/_BAC=/_ADB and color(red)(/_ACB=/_BAD............(1a))`

And the relation of their corresponding sides is

`(AB)/(BC)=(BD)/(AB)=(AD)/(AC)`

From this relation we now consider

`(BD)/(AB)=(AD)/(AC)`

`=>AD*AB=AC*BD`

`=>AE*AB=AC*AC=AC^2,`

`(" since "AD=AE and BD=AC)`

Now rearranging the above relation we can write

`(AE)/(AC)=(AC)/(AB).........................(2)`

This relation (2) reveals that

`DeltaABC and Delta AEC" are similar"`

`"Their corresponding angles are"`

`/_ABC=/_ACE and color(red)( /_ACB =/_CEA .................(2b))`

Comparing Relation (1a) and (2b) we can get

`color(green)(/_BAD=/_CEA`

Proved