Answer: The proof is mentioned below.
Step-by-step explanation:
Here, Δ ABC is isosceles triangle.
Therefore, AB = BC
Prove: Δ ABO ≅ Δ ACO
In Δ ABO and Δ ACO,
∠ BAO ≅ ∠ CAO ( AO bisects ∠ BAC )
∠ AOB ≅ ∠ AOC ( AO is perpendicular to BC )
BO ≅ OC ( O is the mid point of BC)
Thus, By ASA postulate of congruence,
Δ ABO ≅ Δ ACO
Therefore, By CPCTC,
∠B ≅ ∠ C
Where ∠ B and ∠ C are the base angles of Δ ABC.
Step-by-step explanation:
Let the width be x feet.
Therefore, length = 2x - 10
Perimeter of rectangle = 52
Extraneous solutions are the values that we get when solving equations which aren't really solutions to the equation.
<h3>
What are extraneous solutions?</h3>
Your information is incomplete. Therefore, an overview will be given. An extraneous solution is the root of a transformed equation which is not a root of the original equation since it was excluded from the domain of the original equation.
The reason extraneous solutions exist is simply that some operations produce extra answers, and these operations are a part of the path to solving the problem.
Learn more about equations on:
brainly.com/question/2972832
Answer:
thx i ned the points lol
Step-by-step explanation:
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