Respuesta :

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.
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