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.
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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

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,

Answer 2

Since, ∠AID=∠ABD=90°(Given), this means that ADBI is an isosceles quadrilateral. Therefore, CDA is an isosceles triangle with CD=CA. Since, we know that CI bisects ∠C, which means that CI is perpendicular to AD. Hence proved