Answer:
The First step is
Construction:
Construct an altitude to hypotenuse c and label it CD.
The Proof is below.
Step-by-step explanation:
Given:
In triangle ABC, angle C is 90°
AB = c , AC = b , BC = a
Construction:
Construct an altitude to hypotenuse c and label it CD
segment CD is perpendicular to segment AB.
To Prove:
Pythagoras Theorem
[tex](BC)^{2}=(AB)^{2}+(AC)^{2}[/tex]
Proof:
In Δ ABC and Δ CBD
∠C ≅ ∠D …………..{ measure of each angle is 90° given }
∠B ≅ ∠B ……….....{Reflexive Property}
Δ ABC ~ Δ CBD ….{Angle-Angle Similarity test}
If two triangles are similar then their sides are in proportion.
[tex]\dfrac{AB}{CB} =\dfrac{BC}{BD} \textrm{corresponding sides of similar triangles are in proportion}\\[/tex]
By the cross product property we have
[tex](BC)^{2}=AB\times BD[/tex] ............( 1 )
Similarly ,
In Δ ABC and Δ ACD
∠C ≅ ∠D …………..{ measure of each angle is 90° given }
∠A ≅ ∠A ……….....{Reflexive Property}
Δ ABC ~ Δ ACD ….{Angle-Angle Similarity test}
If two triangles are similar then their sides are in proportion.
[tex]\dfrac{AB}{AC} =\dfrac{AC}{AD} \textrm{corresponding sides of similar triangles are in proportion}\\[/tex]
By the cross product property we have
[tex](AC)^{2}=AB\times AD[/tex].......... ( 2 )
Now by Adding 1 and 2 we get
[tex](BC)^{2}+(AC)^{2}=AB\times BD+AB\times AD=AB(BD+DA)\\\\(BC)^{2}+(AC)^{2}=AB(AB)....A-D-B[/tex]
Therefore,
[tex](BC)^{2}+(AC)^{2}=AB^{2}[/tex]
[tex]c^{2}=b^{2}+a^{2}[/tex]
[tex](\textrm{Hypotenuse})^{2} = (\textrm{Shorter leg})^{2}+(\textrm{Longer leg})^{2}[/tex] Pythagoras Theorem......Proved
