Question

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.​

Answer 1

Answer:

Since, ∠AID=∠ABD=90°(Given), this means that ADBI is an isosceles quadrilateral.

⇒∠ADI=∠ABI=45°

⇒∠DAI=45°

But, ∠ADB=∠ADI+∠BDI

=45°+∠IAB

=∠IAD+∠CAI

=∠CAD

Therefore, CDA is an isosceles triangle with CD=CA.

Since, we know that CI bisects ∠C, which means that CI is perpendicular to AD.

Now, BC=a and CA=b, therefore DB=AC-BC=b-a and on using the pythagoras theorem in ΔADB,

AD^{2} =c^{2} +(b-a)^{2}AD

2

=c

2

+(b−a)

2

=c^{2} +b^{2} +a^{2} -2bac

2

+b

2

+a

2

−2ba

=2b(b-a)2b(b−a)

But, 2(ID)^{2}=(AD)^{2} =2b(b-a)2(ID)

2

=(AD)

2

=2b(b−a)

⇒ID=\sqrt{b(b-a)}ID=

b(b−a)

Hence proved.

Answer 2

Answer:

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Explanation:

refer to attachment.. hope it helps