Question

ACD is the exterior angle of ABC. The bisectors of ABC and ACD intersect at de point. Prove that, BEC = 1 2x BAC​

Answer 1

Answer:

$\red{\fbox{Given:-}$ ∆ABC is a triangle where BT and CT are internal and external bisectors of <ABC and <ACB respectively and they meet at T.

$\blue{\fbox{RTP:-}$<BTC = 1/2 <BAC

Construction:- Let, BC is produced to D and the external <ACD is formed.  

$\pink{\fbox{Proof:-}$

In ∆ABC,

Exterior <ACD = interior opposite angle ( <ABC + <BAC )  

=) 1/2 <ACD = 1/2 <ABC+ 1/2 <BAC

=) <TCD = <TBC + 1/2 <BAC ......( 1 )

in ∆TBC, <TCD = <BTC + <TBC ........( 2 )

From ( 1 ) and ( 2 ),

<TBC + 1/2 <BAC = <BTC + <TBC  

=) 1/2 <BAC = <BTC

=) <BTC = 1/2 <BAC

[ PROVED ]

Answer 2

S

Construction:- Let, BC is produced to D and the external <ACD is formed.

Construction:- Let, BC is produced to D and the external <ACD is formed. \pink{\fbox{Proof:-}

Construction:- Let, BC is produced to D and the external <ACD is formed. \pink{\fbox{Proof:-}In ∆ABC,

Ext ( <ABC + <BAC )

( <ABC + <BAC ) =) 1/2 <ACD = 1/2 <ABC+ 1/2 <BAC

( <ABC + <BAC ) =) 1/2 <ACD = 1/2 <ABC+ 1/2 <BAC=) <TCD = <TBC + 1/2 <BAC ......( 1 )

( <ABC + <BAC ) =) 1/2 <ACD = 1/2 <ABC+ 1/2 <BAC=) <TCD = <TBC + 1/2 <BAC ......( 1 )in ∆TBC, <TCD = <BTC + <TBC ........( 2 )

( <ABC + <BAC ) =) 1/2 <ACD = 1/2 <ABC+ 1/2 <BAC=) <TCD = <TBC + 1/2 <BAC ......( 1 )in ∆TBC, <TCD = <BTC + <TBC ........( 2 )From ( 1 ) and ( 2 ),

( <ABC + <BAC ) =) 1/2 <ACD = 1/2 <ABC+ 1/2 <BAC=) <TCD = <TBC + 1/2 <BAC ......( 1 )in ∆TBC, <TCD = <BTC + <TBC ........( 2 )From ( 1 ) and ( 2 ),<TBC + 1/2 <BAC = <BTC + <TBC

( <ABC + <BAC ) =) 1/2 <ACD = 1/2 <ABC+ 1/2 <BAC=) <TCD = <TBC + 1/2 <BAC ......( 1 )in ∆TBC, <TCD = <BTC + <TBC ........( 2 )From ( 1 ) and ( 2 ),<TBC + 1/2 <BAC = <BTC + <TBC =) 1/2 <BAC = <BTC

( <ABC + <BAC ) =) 1/2 <ACD = 1/2 <ABC+ 1/2 <BAC=) <TCD = <TBC + 1/2 <BAC ......( 1 )in ∆TBC, <TCD = <BTC + <TBC ........( 2 )From ( 1 ) and ( 2 ),<TBC + 1/2 <BAC = <BTC + <TBC =) 1/2 <BAC = <BTC=) <BTC = 1/2 <BAC

your answer

∆ ABC =<B +<A ( €x +prop)

1/2 <ACD = 1. /2 <B +1/2 <A

< 1= 1/2 <1/2 <B <A________(1)

in ∆ ABC ,BC is a extended to D

will gate Final Answer as = < BAC = 1/2 BAC

Refer the attachment