Short Pythagorean Theorem Proof

[nycmathguy]

Set 1.4

Can you do parts (a-d)? This will be math notes for me. Thanks.

 

[MathLover1]

a.

in triangle BCD and CDA

given < CAD = < DCB
also
< BCA = < CDB (both 90 degrees)
CD=CD (common side)
=>triangles BCD and BAC are similar
if so, then corresponding sides are proportional
a/y=c/a

b.

a/y=c/a->a^2=cy


c. Show that triangle ACD is similar to ABC , and use this to deduce that b^2=c^2-cy.

triangle ACD is similar to ABC: all angles are congruent
each have 90 degree angle, given < CAD = < DCB, so third angles are equal

then sides are proportional
b/(c-y)=c/b cross multiply
b^2=c(c-y)
b^2=c^2-cy

(d)

We need to combine the two equations in part (b) and (c) to arrive at c^2=a^2+b^2

a^2=cy
b^2=c^2-cy............substitute cy
b^2=c^2-a^2
a^2+b^2=c^2

 

[nycmathguy]

a.

in triangle BCD and CDA

given < CAD = < DCB
also
< BCA = < CDB (both 90 degrees)
CD=CD (common side)
=>triangles BCD and BAC are similar
if so, then corresponding sides are proportional
a/y=c/a

b.

a/y=c/a->a^2=cy


c. Show that triangle ACD is similar to ABC , and use this to deduce that b^2=c^2-cy.

triangle ACD is similar to ABC: all angles are congruent
each have 90 degree angle, given < CAD = < DCB, so third angles are equal

then sides are proportional
b/(c-y)=c/b cross multiply
b^2=c(c-y)
b^2=c^2-cy

(d)

We need to combine the two equations in part (b) and (c) to arrive at c^2=a^2+b^2

a^2=cy
b^2=c^2-cy............substitute cy
b^2=c^2-a^2
a^2+b^2=c^2

Thank you very much.