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The sides AB , AC of a triangle ABC are equal and P is any
point within the triangle on the bisector of angle BAC . BP
produced meets AC in Q . Prove that BP > PQ ?
point within the triangle on the bisector of angle BAC . BP
produced meets AC in Q . Prove that BP > PQ ?
Most Upvoted Answer
The sides AB , AC of a triangle ABC are equal and P is anypoint within...
Proof:
Let's assume that the triangle ABC is isosceles, with AB = AC. We need to prove that BP > PQ, where P lies on the angle bisector of BAC and Q is the intersection point of the extended line BP and AC.
1. Construction:
Construct the angle bisector of BAC and label it as AD, where D is the point of intersection with BC.
Draw a line parallel to AD from point B, intersecting AC at point Q.
2. Proof:
Since AD is the angle bisector of BAC, we can conclude that angles BAD and CAD are equal.
Therefore, triangles ABD and ACD are congruent by the SAS (side-angle-side) congruence criterion.
Hence, AD = AD (common side), AB = AC (given), and angles BAD and CAD are equal.
3. Property 1:
We can observe that triangles ABP and ACQ are similar by the AA (angle-angle) similarity criterion.
Both triangles have a common angle at A, and the angles at B and C are equal because they are corresponding angles when AB is parallel to CQ.
4. Property 2:
From the similarity of triangles ABP and ACQ, we can write the following proportion:
AB / AC = BP / CQ
Since AB = AC (given), the above proportion simplifies to:
1 = BP / CQ
5. Property 3:
Considering the fact that CQ is the extended line segment of BP, we can conclude that CQ > BP.
Hence, 1 = BP / CQ < 1,="" which="" implies="" that="" bp="" />< />
6. Conclusion:
From Property 3, we can say that BP < cq,="" which="" means="" that="" bp="" is="" always="" less="" than="" />
Therefore, we have proved that BP > PQ, as PQ is a part of CQ.
Hence, the statement is true for any point P lying on the angle bisector of BAC in an isosceles triangle ABC with AB = AC.
Let's assume that the triangle ABC is isosceles, with AB = AC. We need to prove that BP > PQ, where P lies on the angle bisector of BAC and Q is the intersection point of the extended line BP and AC.
1. Construction:
Construct the angle bisector of BAC and label it as AD, where D is the point of intersection with BC.
Draw a line parallel to AD from point B, intersecting AC at point Q.
2. Proof:
Since AD is the angle bisector of BAC, we can conclude that angles BAD and CAD are equal.
Therefore, triangles ABD and ACD are congruent by the SAS (side-angle-side) congruence criterion.
Hence, AD = AD (common side), AB = AC (given), and angles BAD and CAD are equal.
3. Property 1:
We can observe that triangles ABP and ACQ are similar by the AA (angle-angle) similarity criterion.
Both triangles have a common angle at A, and the angles at B and C are equal because they are corresponding angles when AB is parallel to CQ.
4. Property 2:
From the similarity of triangles ABP and ACQ, we can write the following proportion:
AB / AC = BP / CQ
Since AB = AC (given), the above proportion simplifies to:
1 = BP / CQ
5. Property 3:
Considering the fact that CQ is the extended line segment of BP, we can conclude that CQ > BP.
Hence, 1 = BP / CQ < 1,="" which="" implies="" that="" bp="" />< />
6. Conclusion:
From Property 3, we can say that BP < cq,="" which="" means="" that="" bp="" is="" always="" less="" than="" />
Therefore, we have proved that BP > PQ, as PQ is a part of CQ.
Hence, the statement is true for any point P lying on the angle bisector of BAC in an isosceles triangle ABC with AB = AC.
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The sides AB , AC of a triangle ABC are equal and P is anypoint within...
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The sides AB , AC of a triangle ABC are equal and P is anypoint within the triangle on the bisector of angle BAC . BPproduced meets AC in Q . Prove that BP > PQ ?
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The sides AB , AC of a triangle ABC are equal and P is anypoint within the triangle on the bisector of angle BAC . BPproduced meets AC in Q . Prove that BP > PQ ? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about The sides AB , AC of a triangle ABC are equal and P is anypoint within the triangle on the bisector of angle BAC . BPproduced meets AC in Q . Prove that BP > PQ ? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sides AB , AC of a triangle ABC are equal and P is anypoint within the triangle on the bisector of angle BAC . BPproduced meets AC in Q . Prove that BP > PQ ?.
The sides AB , AC of a triangle ABC are equal and P is anypoint within the triangle on the bisector of angle BAC . BPproduced meets AC in Q . Prove that BP > PQ ? for Class 9 2024 is part of Class 9 preparation. The Question and answers have been prepared according to the Class 9 exam syllabus. Information about The sides AB , AC of a triangle ABC are equal and P is anypoint within the triangle on the bisector of angle BAC . BPproduced meets AC in Q . Prove that BP > PQ ? covers all topics & solutions for Class 9 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for The sides AB , AC of a triangle ABC are equal and P is anypoint within the triangle on the bisector of angle BAC . BPproduced meets AC in Q . Prove that BP > PQ ?.
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