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Given: ΔABC is a right triangle.

Prove: a2 + b2 = c2


Right triangle BCA with sides of length a, b, and c. Perpendicular CD forms right triangles BDC and CDA. CD measures h units, BD measures y units, DA measures x units.


The two-column proof with missing justifications proves the Pythagorean Theorem using similar triangles:


Which is not a justification for the proof?



Substitution


Addition Property of Equality


Transitive Property of Equality


Distributive Property of Equality

Given: ΔABC is a right triangle. Prove: a2 + b2 = c2 Right triangle BCA with sides of length a, b, and c. Perpendicular CD forms right triangles BDC and CDA. CD class=

Verified Answer :

Nirina7

the Pythagorean Theorem
proof of
let ΔABC be
a right triangle.
and sinA=a/c, and cosA= b/c
a opposite side of the angle A
b the adjacent side of the angle A
and c is the hypotenus

we know that  sin²A +cos²A= (a/c)²+ (b/c) ², but  sin²A +cos²A=1
so,  a²/c²+ b²/c ²=1 which implies  a²+ b²=c²


the answer is

Transitive Property of Equality

proof

the right triangles BDC and CDA are similars



MissPhiladelphia

We start with the original right triangle, now denoted ABC, and need only one additional construct - the altitude AD. The triangles ABC, DBA, and DAC are similar which leads to two ratios:

AB/BC = BD/AB and AC/BC = DC/AC.

Written another way these become

AB·AB = BD·BC and AC·AC = DC·BC

Summing up we get

AB·AB + AC·AC= BD·BC + DC·BC 
                         = (BD+DC)·BC = BC·BC.
so not in the proof is Transitive Property of Equality

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