The problem statement tells us that
- AM ≅ MB, ∴ ΔAMB is isosceles
- AM ≅ MC, ∴ ΔAMC is isosceles
Base angles of an isosceles triangle are equal, so ∠MAB ≅ ∠MBA. ∠AMC is the exterior angle opposite those two, so it is equal to their sum, 2∠MAB.
The base angles in isosceles ΔAMC are equal to half the difference between the apex angle, ∠AMC, and 180°. That is,
... ∠MAC = (1/2)(180° -∠AMC) = (1/2)(180° -2∠MAB) = 90° -∠MAB
The angle at A of ΔABC is the sum of the two angles created by the median AM. That is ...
∠A = ∠MAC + ∠MAB
∠A = (90° -∠MAB) +∠MAB
∠A = 90°
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Maybe a shorter way to get there is to realize that ...
... MA ≅ MB ≅ MC
so M is the center of a circle with BC as a diameter and A a point on the circle. Angle BAC is inscribed in the semicircle and subtends an arc of 180°, so angle A is 90°.
Answer:
1. vertex
2. axis of symmetry
3. I'm not sure
4. x-intercept
5. y-intercept
Step-by-step explanation:
sorry i knew the other 4, just not #3.
I hope this helps
Answer:
Standard form:
3n-9
Step-by-step explanation:
I hope this helps! Have a nice day!
Problem 7)
The sides 2x+4 and 3x-10 are congruent because they are opposite congruent angles. This is an isosceles triangle. Since they are congruent, they are equal. Let's solve for x.
2x+4 = 3x-10
2x+4-2x = 3x-10-2x
4 = x-10
4+10 = x-10+10
14 = x
x = 14
The answer to problem 7 is x = 14
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Problem 8)
This is a right triangle because of the square angle marker. That angle is 90 degrees. This is also an isosceles triangle because of the similar tick marks on the two upper sides. Because this is an isosceles right triangle, that means the triangle is a 45-45-90 triangle. Two angles are 45 degrees.
So,
2x+3 = 45
2x+3-3 = 45-3
2x = 42
2x/2 = 42/2
x = 21
The final answer to problem 8 is x = 21
Answer:
55°
Step-by-step explanation:
∠KJL=180° - 125 = 55°
KL=KJ
so, ∠KJL= ∠KLJ =55°
