Question

justin is using the figure shown below to prove Pythagorean Theorem using triangle similarity: In the given triangle ABC, angle A is 90o and segment AD is perpendicular to segment BC. Which of these could be a step to prove that BC2 = BA2 + CA2? A. By segment addition, DC plus AD is equal to BC. B. By segment addition, DC plus AD is equal to AB. C.By segment addition, DC plus BD is equal to BC. D. By segment addition, DC plus BD is equal to AB.

Answer 1

Answer: C.By segment addition, DC plus BD is equal to BC.

Step-by-step explanation:

Here, ABC, angle A is 90° and segment AD is perpendicular to segment BC.

In triangles ADB and ABC,

$m\angle ADB=m\angle CAB$ (right angles)

$m\angle B=m\angle B$( reflexive angles)

Therefore,  $\triangle ADB\sim \triangle ABC$

Therefore, BA/BD =BC/BA (By the property of similar triangles)

⇒$BA^2=BC.BD$ ------- (1)

Similarly, $\triangle ADC\sim \triangle ABC$

⇒$CA^2=BC.CD$--------(2)

After adding equation (1) and (2)

We get $BA^2+CA^2= BC.BD+ BC.CD$

$BA^2+CA^2= BC( BD+CD)$

$BA^2+CA^2=BC.BC$( By segment addition, DC plus BD is equal to BC which is shown in figure)

$BA^2+CA^2=BC^2$

Answer 2

The answer to this question is letter C) By segment addition, DC plus BD is equal to BC. Point D is located at the middle of segment BC. Therefore BC is just equal to BD plus DC.

Now we can prove BC2 = BA2+CA2 by using the Pythagorean theorem.