are congruent.28, in the figure BA·LAC and DE.L EF such that BA-DE an

1.	are congruent.28, in the figure BA·LAC and DE.L EF such that BA-DE and BF= DC. Prove tr29. From the vertices RndAROhat AC EFCofAARn
Answer» Given: ABAC and DEFE such that, AB = DE and BF = CD To prove: AC = EF Proof: InABC, we have, BC = BF + FC and, inDEF FD = FC + CD But, BF = CD [Given] So, BC = BF + FC and, FD = FC + BF BC = FD So, inABC andDEF, we have, BAC =DEF = 90o[Given] BC = FD [Proved above] AB = DE [Given] Thus, by Right angle-Hypotenuse-Side criterion of congruence, we have ABCDEF 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 BCDBBAE The corresponding parts of the congruent triangles are equal. So, CD = AE [Proved]

are congruent.28, in the figure BA·LAC and DE.L EF such that BA-DE and BF= DC. Prove tr29. From the vertices RndAROhat AC EFCofAARn

Answer»

Given: ABAC and DEFE such that,

AB = DE and BF = CD

To prove: AC = EF

Proof:

InABC, we have,

BC = BF + FC

and, inDEF

FD = FC + CD

But, BF = CD [Given]

So, BC = BF + FC

and, FD = FC + BF

BC = FD

So, inABC andDEF, we have,

BAC =DEF = 90o[Given]

BC = FD [Proved above]

AB = DE [Given]

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

ABCDEF

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

BCDBBAE

The corresponding parts of the congruent triangles are equal.

So, CD = AE [Proved]