Hypotenuse Leg Theorem

Consider $△ A B C$ and $△ D E F,$ shown below.

By applying the Pythagorean Theorem in each triangle, the following equations can be written.

{c^{2} = a^{2} + b_{1}^{2} c^{2} = a^{2} + b_{2}^{2} (I) (II)

The expression on the right hand-side of the first equation can be substituted into the second equation. Then a relation between

b_{1}

and

b_{2}

can be found.

{c^{2} = a^{2} + b_{1}^{2} c^{2} = a^{2} + b_{2}^{2} (I) (II)

Substitute

$(II):$ $c^{2} = a^{2} + b_{1}^{2}$

{c^{2} = a^{2} + b_{1}^{2} a^{2} + b_{1}^{2} = a^{2} + b_{2}^{2}

Solve for

b_{1}

SubEqn

$(II):$ $LHS - a^{2} = RHS - a^{2}$

{c^{2} = a^{2} + b_{1}^{2} b_{1}^{2} = b_{2}^{2}

SqrtEqn

$(II):$ $LHS = RHS$

{c^{2} = a^{2} + b_{1}^{2} b_{1}^{2} = b_{2}^{2}

SqrtToAbs

$(II):$ $a^{2} = ∣ a ∣$

{c^{2} = a^{2} + b_{1}^{2} ∣ b_{1} ∣ = ∣ b_{2} ∣

Since both

b_{1}

and

b_{2}

represent side lengths, they are positive numbers. Moreover, the absolute value of a positive number is the number itself. Therefore, the second equation implies that

b_{1}

and

b_{2}

are equal.

∣ b_{1} ∣ b_{1} b_{2} = ∣ b_{2} ∣ > 0 > 0 \Rightarrow b_{1} = b_{2}

Consequently, the three sides of

△ A B C

are congruent to the corresponding three sides of

△ D E F .

Therefore, by the Side-Side-Side Congruence Theorem the triangles are congruent.

$△ A B C ≅ △ D E F$

{{ article.displayTitle }}

Hypotenuse Leg Theorem

Proof

	{{ 'ml-lesson-number-slides' \| message : article.intro.bblockCount}}
	{{ 'ml-lesson-number-exercises' \| message : article.intro.exerciseCount}}
	{{ 'ml-lesson-time-estimation' \| message }}