Hence they are similar
#2
- <A=<H
- And ratio of sides are same
Hence both triangles are similar by SAS property
#3
Answer:
x° = ∠OBR = ∠ABC (base angles of a cyclic isosceles trapezoid)
Step-by-step explanation:
APRB form a cyclic trapezoid
∠APO = x° (Base angle of an isosceles triangle)
∠OPR = ∠ORP (Base angle of an isosceles triangle)
∠ORB = ∠OBR (Base angle of an isosceles triangle)
∠APO + ∠OPR + ∠OBR = 180° (Sum of opposite angles in a cyclic quadrilateral)
Similarly;
∠ORB + ∠ORP + x° = 180°
Since ∠APO = x° ∠ORB = ∠OBR and ∠OPR = ∠ORP we put
We also have;
∠OPR = ∠AOP = ∠BOR (Alternate interior angles of parallel lines)
Hence 2·x° + ∠AOP = 180° (Sum of angles in a triangle) = 2·∠OBR + ∠BOR
Therefore, 2·x° = 2·∠OBR, x° = ∠OBR = ∠ABC.
Answer:
35
Step-by-step explanation:
