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In the adjoining quardilateral ABCD , AO and BO are the bisectors of angle A and angle B respectively . Prove that angle AOB = 1/ 2 ( angle c + angle D?
( Please answer fast it's urgent please pls pls pls)?
( Please answer fast it's urgent please pls pls pls)?
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In the adjoining quardilateral ABCD , AO and BO are the bisectors of a...
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In the adjoining quardilateral ABCD , AO and BO are the bisectors of a...
Proof:
Given: Quadrilateral ABCD, where AO and BO are the bisectors of ∠A and ∠B respectively.
To prove: ∠AOB = 1/2 (∠C + ∠D)
Construction:
Draw the bisectors OC and OD of ∠C and ∠D respectively.
Proof:
Step 1:
Since AO is the bisector of ∠A, we have ∠OAC = ∠OAB (Angle bisector theorem).
Step 2:
Similarly, since BO is the bisector of ∠B, we have ∠OBD = ∠OBA (Angle bisector theorem).
Step 3:
Adding ∠OAC and ∠OBD, we get ∠OAC + ∠OBD = ∠OAB + ∠OBA.
Step 4:
Since ∠OAC = ∠OAB and ∠OBD = ∠OBA, we can rewrite the equation as 2∠OAC = 2∠OBA.
Step 5:
Dividing both sides of the equation by 2, we have ∠OAC = ∠OBA.
Step 6:
Since ∠OAC = ∠OBA, we can conclude that triangle AOC is congruent to triangle BOA by the angle-side-angle (ASA) congruence criterion.
Step 7:
Therefore, AO = BO (corresponding parts of congruent triangles are congruent).
Step 8:
Since AO = BO, triangle AOB is an isosceles triangle.
Step 9:
In an isosceles triangle, the angles opposite to the equal sides are also equal.
Step 10:
Therefore, ∠AOB = ∠OAB + ∠OBA = ∠OAC + ∠OBD.
Step 11:
Now, consider triangle COD.
Step 12:
Since OC is the bisector of ∠C, we have ∠OCA = ∠OCB (Angle bisector theorem).
Step 13:
Similarly, since OD is the bisector of ∠D, we have ∠ODC = ∠ODB (Angle bisector theorem).
Step 14:
Adding ∠OCA and ∠ODC, we get ∠OCA + ∠ODC = ∠OCB + ∠ODB.
Step 15:
Since ∠OCA = ∠OCB and ∠ODC = ∠ODB, we can rewrite the equation as 2∠OCA = 2∠OCB.
Step 16:
Dividing both sides of the equation by 2, we have ∠OCA = ∠OCB.
Step 17:
Since ∠OCA = ∠OCB, we can conclude that triangle COA is congruent to triangle COB by the angle-side-angle (ASA) congruence criterion.
Step 18:
Therefore, OA = OB (
Given: Quadrilateral ABCD, where AO and BO are the bisectors of ∠A and ∠B respectively.
To prove: ∠AOB = 1/2 (∠C + ∠D)
Construction:
Draw the bisectors OC and OD of ∠C and ∠D respectively.
Proof:
Step 1:
Since AO is the bisector of ∠A, we have ∠OAC = ∠OAB (Angle bisector theorem).
Step 2:
Similarly, since BO is the bisector of ∠B, we have ∠OBD = ∠OBA (Angle bisector theorem).
Step 3:
Adding ∠OAC and ∠OBD, we get ∠OAC + ∠OBD = ∠OAB + ∠OBA.
Step 4:
Since ∠OAC = ∠OAB and ∠OBD = ∠OBA, we can rewrite the equation as 2∠OAC = 2∠OBA.
Step 5:
Dividing both sides of the equation by 2, we have ∠OAC = ∠OBA.
Step 6:
Since ∠OAC = ∠OBA, we can conclude that triangle AOC is congruent to triangle BOA by the angle-side-angle (ASA) congruence criterion.
Step 7:
Therefore, AO = BO (corresponding parts of congruent triangles are congruent).
Step 8:
Since AO = BO, triangle AOB is an isosceles triangle.
Step 9:
In an isosceles triangle, the angles opposite to the equal sides are also equal.
Step 10:
Therefore, ∠AOB = ∠OAB + ∠OBA = ∠OAC + ∠OBD.
Step 11:
Now, consider triangle COD.
Step 12:
Since OC is the bisector of ∠C, we have ∠OCA = ∠OCB (Angle bisector theorem).
Step 13:
Similarly, since OD is the bisector of ∠D, we have ∠ODC = ∠ODB (Angle bisector theorem).
Step 14:
Adding ∠OCA and ∠ODC, we get ∠OCA + ∠ODC = ∠OCB + ∠ODB.
Step 15:
Since ∠OCA = ∠OCB and ∠ODC = ∠ODB, we can rewrite the equation as 2∠OCA = 2∠OCB.
Step 16:
Dividing both sides of the equation by 2, we have ∠OCA = ∠OCB.
Step 17:
Since ∠OCA = ∠OCB, we can conclude that triangle COA is congruent to triangle COB by the angle-side-angle (ASA) congruence criterion.
Step 18:
Therefore, OA = OB (
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In the adjoining quardilateral ABCD , AO and BO are the bisectors of angle A and angle B respectively . Prove that angle AOB = 1/ 2 ( angle c + angle D?( Please answer fast it's urgent please pls pls pls)?
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In the adjoining quardilateral ABCD , AO and BO are the bisectors of angle A and angle B respectively . Prove that angle AOB = 1/ 2 ( angle c + angle D?( Please answer fast it's urgent please pls pls pls)? for Class 8 2024 is part of Class 8 preparation. The Question and answers have been prepared according to the Class 8 exam syllabus. Information about In the adjoining quardilateral ABCD , AO and BO are the bisectors of angle A and angle B respectively . Prove that angle AOB = 1/ 2 ( angle c + angle D?( Please answer fast it's urgent please pls pls pls)? covers all topics & solutions for Class 8 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the adjoining quardilateral ABCD , AO and BO are the bisectors of angle A and angle B respectively . Prove that angle AOB = 1/ 2 ( angle c + angle D?( Please answer fast it's urgent please pls pls pls)?.
In the adjoining quardilateral ABCD , AO and BO are the bisectors of angle A and angle B respectively . Prove that angle AOB = 1/ 2 ( angle c + angle D?( Please answer fast it's urgent please pls pls pls)? for Class 8 2024 is part of Class 8 preparation. The Question and answers have been prepared according to the Class 8 exam syllabus. Information about In the adjoining quardilateral ABCD , AO and BO are the bisectors of angle A and angle B respectively . Prove that angle AOB = 1/ 2 ( angle c + angle D?( Please answer fast it's urgent please pls pls pls)? covers all topics & solutions for Class 8 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the adjoining quardilateral ABCD , AO and BO are the bisectors of angle A and angle B respectively . Prove that angle AOB = 1/ 2 ( angle c + angle D?( Please answer fast it's urgent please pls pls pls)?.
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