Question

In the following, Triangle FEC congruent Triangle GBD and angle 1= angle 2. prove that Triangle ADE similar triangle ABC.

Answer 1

HELLO DEAR,

Given : ∆ FEC ≅ ∆ GBD  ,

SO From CPCT

WE GET,

BD = CE ----------------- ( 1 )

ALSO GIVEN : ∠ 1 =  ∠ 2  ,

SO FROM BASE

ANGLE THEOREM IN ∆ ADE

WE GET

AD  =AE ------------------------ ( 2 ) 

From equation 1 and 2 we get

ADBD = AECE , So from converse of B.P.T. we get

DE | | BC

THEN ,

∠ 1  = ∠ 3                                 
( CORRESPONDING ANGLES AS DE  | | BC AND AB IS TRANSVERSAL LINE )

AND

∠ 2  = ∠ 4                                  ( CORRESPONDING ANGLES AS DE  | | BC AND AC IS TRANSVERSAL LINE )

FROM ABOVE TWO EQUATIONS WE CAN SAY THAT :

∆ ADE ~ ∆ ABC        ( ByAArule )
( Hence proved )



I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

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