1. [tex] m\angle BAC=m\angle CAD,\ m\angle ACB=m\angle ADC=90^{\circ} [/tex], then [tex] m\angle ABC=m\angle ACD [/tex] and triangles ADC and ACB are similar by AAA theorem.
2. The ratio of the corresponding sides of similar triangles is constant, so
[tex] \dfrac{AC}{AB}= \dfrac{AD}{AC} [/tex].
3. Knowing lengths you could state that [tex] \dfrac{b}{c}= \dfrac{e}{b} [/tex].
4. This ratio is equivalent to [tex] b^2=ce [/tex].
5. [tex] m\angle ABC=m\angle CBD,\ m\angle ACB=m\angle CDB=90^{\circ} [/tex], then [tex] m\angle BAC=m\angle BCD [/tex] and triangles BDC and BCA are similar by AAA theorem.
6. The ratio of the corresponding sides of similar triangles is constant, so
[tex] \dfrac{BC}{BD}= \dfrac{AB}{BC} [/tex].
7. Knowing lengths you could state that [tex] \dfrac{a}{d}= \dfrac{c}{a} [/tex].
8. This ratio is equivalent to [tex] a^2=cd [/tex].
9. Now add results of parts 4 and 8:
[tex] b^2+a^2=ce+cd [/tex].
10. c is common factor, then:
[tex] b^2+a^2=c(e+d) [/tex].
11. Since [tex] e+d=c [/tex] you have [tex] a^2+b^2=c\cdot c=c^2 [/tex].
Answer:
Given information: [tex]\triangle ADC\sim \triangle ACB[/tex] and [tex]\triangle BDC\sim \triangle BCA[/tex].
To Prove : [tex]a^2+b^2=c^2[/tex]
Proof:
Statement Proof
1. [tex]\triangle ADC\sim \triangle ACB[/tex] 1. Given
2. [tex]\dfrac{AC}{AB}=\dfrac{AD}{AC}[/tex] 2. The ratio of corresponding parts of similar triangles is a constant
3. [tex]\dfrac{b}{c}=\dfrac{e}{b}[/tex] 3. Rewrite statement 2 using given side labels.
4. [tex]b^2=ce[/tex] 4. Cross multiply statement 4
5. [tex]\triangle BDC\sim \triangle BCA[/tex] 5. Given
6. [tex]\dfrac{BC}{BA}=\dfrac{BD}{BC}[/tex] 6. The ratio of corresponding parts of similar triangles is a constant
7. [tex]\dfrac{a}{c}=\dfrac{d}{a}[/tex] 7. Rewrite statement 5 using given side labels.
8. [tex]a^2=cd[/tex] 8. Cross multiply statement 7
9. [tex]a^2+b^2=cd+ce[/tex] 9. Add statement 4 and 8
10. [tex]a^2+b^2=c(d+e)[/tex] 10. c is common factor
11. [tex]a^2+b^2=c^2[/tex] 11. Segment addition property.
Hence proved.
