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

        =c^2+b^2+a^2-2ba

        =2b(b-a)

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

⇒ID=$\sqrt{b} (b-a)$

Hence, Proved.

Explanation:

Hope It Helps you dear Zehrille❤

Answer 2

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=

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

        =c^2+b^2+a^2-2ba

        =2b(b-a)

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

⇒ID=\sqrt{b} (b-a)b(b−a)

Hence, Proved.