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 ?

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Answer 1 · 2016-08-20T15:33:55

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