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.

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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−a2

=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:

this is your answer

Explanation:

First observe that ADBI is a cyclic

quadrilateral since ∠AID = ∠ABD = 90◦

. Hence

∠ADI = ∠ABI = 45◦

. Hence ∠DAI = 45◦

. But

we also have

∠ADB = ∠ADI + ∠IDB = 45◦ + ∠IAB

= ∠DAI + ∠IAC = ∠DAC.

Therefore CDA is an isosceles triangle with CD =

CA. Since CI bisects ∠C it follows that CI ⊥ AD.

This shows that DB = CA − CB = b − a. Therefore

AD2 = c

2 + (b − a)

2 = c

2 + b

2 + a

2 − 2ba = 2b(b − a).

But then 2ID2 = AD2 = 2b(b − a) and this gives ID =

p

b(b − a).