In ABD& BED ABD = DBE (BD is the angle bisector) ADB = DEB = 90^∘ (given) By AA , ABD∼ DBE By CPST ADBA = DEBD = k(let) & BDBA = BEBD = l(let) AD .
I am not good ⊂((・▽・))⊃
come na please (。•́︿•̀。)
Given : In
Δ
A
B
C
,
=
BD and CE are the bisectors of
∠
a
n
d
r
e
s
p
c
t
i
v
l
y
To prove : BD = CE
Proof : In
∴
(Angles opposite to equal sides)
1
2
D
E
Now, in
BC = BC (Common)
(Equal angles)
(Proved)
≅
(ASA axiom)
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Answers & Comments
In ABD& BED ABD = DBE (BD is the angle bisector) ADB = DEB = 90^∘ (given) By AA , ABD∼ DBE By CPST ADBA = DEBD = k(let) & BDBA = BEBD = l(let) AD .
I am not good ⊂((・▽・))⊃
come na please (。•́︿•̀。)
Given : In
Δ
A
B
C
,
A
B
=
A
C
BD and CE are the bisectors of
∠
B
a
n
d
∠
C
r
e
s
p
e
c
t
i
v
e
l
y
To prove : BD = CE
Proof : In
Δ
A
B
C
,
A
B
=
A
C
∴
∠
B
=
∠
C
(Angles opposite to equal sides)
∴
1
2
∠
B
=
1
2
∠
C
∠
D
B
C
=
∠
E
C
B
Now, in
Δ
D
B
C
a
n
d
Δ
E
B
C
,
BC = BC (Common)
∠
C
=
∠
B
(Equal angles)
∠
D
B
C
=
∠
E
C
B
(Proved)
∴
Δ
D
B
C
≅
E
B
C
(ASA axiom)
∴
B
D
=
C
E