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Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.?🙏?
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Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre...
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Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre...
Proof:
Step 1: Prove CI is perpendicular to AD
Let E be the point of intersection of AI and BC. Then, we have AE = EC (since I is the incenter of ABC).
Also, we have IB = IC (since I is the incenter of ABC and ∠B = 90◦). Therefore, we have ∠IBC = ∠ICB = 1/2(∠B) = 45◦.
Hence, we have ∠IBD = 90◦ - ∠IBC = 45◦.
Now, in △AID, we have ∠AID = 90◦ (since ID is perpendicular to AI).
Therefore, we have ∠DIA = 180◦ - ∠AID - ∠IDA = 180◦ - 90◦ - ∠IDA = 90◦ - ∠IDA.
Also, we have ∠EIC = ∠IBC = 45◦.
Therefore, we have ∠DIC = ∠EIC - ∠EID = 45◦ - ∠EID.
Since AE = EC, we have ∠EID = ∠DIA/2 (since I is the incenter of ABC).
Thus, we have ∠DIC = 45◦ - ∠DIA/2.
Hence, we have ∠DIA + ∠DIC = 90◦.
Therefore, we have CI is perpendicular to AD (since the sum of angles in △DIC is 180◦).
Step 2: Prove ID = sqrt(b(b-a))
In △AID, we have ∠AID = 90◦, and AI is the angle bisector of ∠BAC.
Therefore, we have AD/AB = ID/IB = (b - ID)/BC.
Simplifying, we get AD = b(b - ID)/a.
Hence, we have ID = a(b - AD)/b.
Substituting AD = b(b - ID)/a, we get ID = sqrt(b(b - a)).
Therefore, we have proved that CI is perpendicular to AD, and ID = sqrt(b(b - a)).
Step 1: Prove CI is perpendicular to AD
Let E be the point of intersection of AI and BC. Then, we have AE = EC (since I is the incenter of ABC).
Also, we have IB = IC (since I is the incenter of ABC and ∠B = 90◦). Therefore, we have ∠IBC = ∠ICB = 1/2(∠B) = 45◦.
Hence, we have ∠IBD = 90◦ - ∠IBC = 45◦.
Now, in △AID, we have ∠AID = 90◦ (since ID is perpendicular to AI).
Therefore, we have ∠DIA = 180◦ - ∠AID - ∠IDA = 180◦ - 90◦ - ∠IDA = 90◦ - ∠IDA.
Also, we have ∠EIC = ∠IBC = 45◦.
Therefore, we have ∠DIC = ∠EIC - ∠EID = 45◦ - ∠EID.
Since AE = EC, we have ∠EID = ∠DIA/2 (since I is the incenter of ABC).
Thus, we have ∠DIC = 45◦ - ∠DIA/2.
Hence, we have ∠DIA + ∠DIC = 90◦.
Therefore, we have CI is perpendicular to AD (since the sum of angles in △DIC is 180◦).
Step 2: Prove ID = sqrt(b(b-a))
In △AID, we have ∠AID = 90◦, and AI is the angle bisector of ∠BAC.
Therefore, we have AD/AB = ID/IB = (b - ID)/BC.
Simplifying, we get AD = b(b - ID)/a.
Hence, we have ID = a(b - AD)/b.
Substituting AD = b(b - ID)/a, we get ID = sqrt(b(b - a)).
Therefore, we have proved that CI is perpendicular to AD, and ID = sqrt(b(b - a)).
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Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.?🙏?
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Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.?🙏? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.?🙏? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.?🙏?.
Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.?🙏? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.?🙏? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.?🙏?.
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Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.?🙏?, a detailed solution for Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.?🙏? has been provided alongside types of Let ABC be a right-angled triangle with ∠B = 90◦.Let I be the incentre of ABC. Draw a line perpendicular to AI at I. Let it intersect the line CB at D. Prove that CI is perpendicular to AD and prove that ID =Sqrt(b(b − a)) ,where BC = a and CA = b.?🙏? theory, EduRev gives you an
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