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 ?

1 Answer
Aug 20, 2016

drawn

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