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1 year ago

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In ABC, side BC is extended to point E. When connected to vertex A, segment EA is parallel to segment BD. In this construction,

you are given that BD bisects

1 answer:

Anuta_ua [19.1K]1 year ago

6 0

Two triangles are said to be <u>congruent</u> if they have <em>similar</em> properties. Thus the required <u>options</u> to complete the <em>paragraph proof</em> are:

a. angle 1 is <u>congruent</u> to angle 2.

b. <em>alternate</em> angles are <u>congruent</u> if two parallel lines are cut by a <em>transversal</em>.

c. $\frac{AD}{CD}$ = $\frac{AB}{CB}$

The <em>similarity property</em> of two or more shapes implies that the <u>shapes</u> are congruent. Thus they have the <em>same</em> properties.

From the given <u>diagram</u> in the question, it can be deduced that

ΔABC ≅ ΔABE (<em>substitution</em> property of equality)

Given that EA is <u>parallel</u> to BD, then:

i. <2 ≅ <3 (<em>corresponding</em> angle property)

ii. <1 ≅ < 4 (<em>alternate</em> angle property)

Thus, the required options to complete the <em>paragraph proof</em> are:

Angle 1 is <em>congruent</em> to angle 2.
Alternate angles are <u>congruent</u> if two parallel lines are cut by a <em>transversal</em>.
$\frac{AD}{CD}$ = $\frac{AB}{CB}$

For more clarifications on the properties of congruent triangles, visit: brainly.com/question/1619927

#SPJ1

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Answer:

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Step-by-step explanation:

I will solve this question with the attached triangle

From the attachment, we have:

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Required

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$\frac{24}{\sin(B)} = \frac{29}{\sin(82^o)}$

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Collect like terms

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