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.
Reduce the expression, if possible, by cancelling the common factors.
Exact Form:
−4/15
Decimal Form:
−0.26
Basically do
y2-y1 divided by x2-x1
14-10 divided by -7-8
which equals -4/15