Question

in the given figure BA is perpendicular to AC and DE is perpendicular to EF such that BA is equal to D E and BF is equal to DC prove that AC is equal to EF<br />

Answer 1

Given: AB  AC and DE  FE such that,

AB = DE and BF = CD

To prove: AC = EF

Proof:

In ABC, we have,

BC = BF + FC

and, in DEF

FD = FC + CD

But, BF = CD [Given]

So, BC = BF + FC

and, FD = FC + BF

 BC = FD

So, in ABC and DEF, we have,

BAC = DEF = 90o [Given]

BC = FD [Proved above]

AB = DE [Given]

Thus, by Right angle-Hypotenuse-Side criterion of congruence, we have

ABC DEF

The corresponding parts of the congruent triangle are equal.

So, AC = EF [c.p.c.t]

AE = BCD [Proved above]

Thus by Angle-Side-Angle criterion of congruence, we have

BCD BBAE

The corresponding parts of the congruent triangles are equal.

So, CD = AE [Proved]

Answer 2

Answer:

AB=DE

BF=CD

to prove AC=EF

In triangle ABC we have

BC=BF+FC

and in DEF

FD=FC+CD

but BF=CD(given)

BC=BF+FC

and FD=FC+BF

BC=FD

BAC=DEF=90°(given)

BC=FD( proved above)

AB=DE(given)

we have ABC DEF

the corresponding triangle are equal

so, AC=EF ( CPCT)