3. Reflect right triangle ABC across line BC. Classify triangle ACA' according to its side lengths. Explain how
you know.

Triangle ACA' has two equal sides, based on the side lengths, triangle ACA' can be classified as: an isosceles triangle.

Recall:

Given that triangle ABC is a right triangle, if we reflect it across line BC, we will have  mirror figure of right triangle ABC, forming triangle ACA'.

That is, two right triangles of ABC, which are reflections of each other, will form triangle ACA'. (See image attached).

Thus, this means that:

Therefore,

Also, CA = C'A (congruent sides)

This means that, triangle ACA' has two equal sides that are opposite to each other.

Therefore, based on the side lengths, triangle ACA' can be classified as: an isosceles triangle.

Learn more here:

link

Isosceles triangle

Step-by-step explanation:

Given

Triangle ABC

Required

Reflect across BC and name the triangle according to length of its side

When the triangle is reflected across line BC, the new triangle is ACA (see new attachment)

And the sides of this ACA are:

AC, CA and AA

Such that

AC = CA

and

From the description above, we have two sides (AC and CA) of the triangle to be equal.

By classification of sides, when two sides of a triangle are equal; such triangle is an isosceles triangle

i think it is 70

step-by-step explanation:

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°