Question

Prove the Converse of the Pythagorean Theorem using similar triangles. The Converse of the Pythagorean Theorem states that when the sum of the squares of the lengths of the legs of the triangle equals the squared length of the hypotenuse, the triangle is a right triangle. Be sure to create and name the appropriate geometric figures.

Answer 1

EF = BC = a ÐF is a right angle. FD = CA = b triangle EF = BC = a angle F is a right angle. FD = CA = b In triangle DEF, By Pythagoras Theorem, a2 + b2 = c2 the given AB=c= a^2 + b^2 square root Theorefore AB = DE But by construction, BC = EF and CA = FD triangle ABC congruent to DEF (S.S.S)
hope it helps

Answer 2

Answer:

To prove the Converse of Pythagorean Theorem:

Step-by-step explanation:

1. Draw a right triangle (Triangle ABC)

2. Draw an altitude from and A to the hypotenuse creating 2 new triangles

3. The part that intersects the hypotenuse will be point D, this makes a right angle

Triangle DBA and triangle ABC both have right angles and share angle B, this means that triangle ABC and triangle DBA are congruent by the AA postulate. Since triangle DAC and triangle ABC both have right angles(angle D and angle A) and they share angle C, triangle DAC and ABC are congruent by the AA postulate as well.

The ratios can be set up as:

AC/BC = DC/AC & AB/BC = DB/BA

Once cross multiplied, the result should be:

(AC)² = (BC)(DC) & (AB)² = (BC)(DB)

Now add both sides together to get:

(AC)² + (AB)² = BC(DC) + (BC)(DB)

To make things easier, factor out the like terms:

(AC)² + (AB)² = BC(DC + DB)

When looking at the triangle(must draw to see), it shows that (DC + DB) is actually side BC. This leaves:

(AC)² + (AB)² = BC(BC) → (AC)² + (AB)² = (BC)²

The Pythagorean theorem says that a² + b² = c². Knowing this, on the original triangle, side AB = a, side AC = b, and side BC = c. Knowing this:

(AC)² + (AB)² = (BC)² → a² + b² = c²

This proves the Converse of the Pythagorean Theorem.