I can’t understand why everyone complicates this question. It can be easily solved by similar triangles.
In this png, we have something to make sure.
∠B=∠DAB
∠
B
=
∠
D
A
B
(Yes, dab)
This also means AD=BD
A
D
=
B
D
.
This is our basic construction of D, which is going to help us.
∠ADC=∠DAB+∠B=2∠B=∠CAB
∠
A
D
C
=
∠
D
A
B
+
∠
B
=
2
∠
B
=
∠
C
A
B
∠CAD=∠CAB−∠DAB=∠B
∠
C
A
D
=
∠
C
A
B
−
∠
D
A
B
=
∠
B
These are based on the fact that ∠A=2×∠B
∠
A
=
2
×
∠
B
Actually these conditions suffice. Because I am just proving that △ACD∼△BCA
△
A
C
D
∼
△
B
C
A
Similarity makes us realize the following:
ACBC=ADAB
A
C
B
C
=
A
D
A
B
and
ACBC=CDAC
A
C
B
C
=
C
D
A
C
So
AC×AB=BC×AD
A
C
×
A
B
=
B
C
×
A
D
and
AC2=BC×CD
A
C
2
=
B
C
×
C
D
So
BC2=BC×(BD+CD)=BC×(AD+CD)
B
C
2
=
B
C
×
(
B
D
+
C
D
)
=
B
C
×
(
A
D
+
C
D
)
=AC×AB+AC2
=
A
C
×
A
B
+
A
C
2
Q.E.D.
2.4k Views ·