Math, asked by Rajam7398, 1 year ago

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.

Answers

Answered by boffeemadrid

Answer:

Step-by-step explanation:

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.

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adityasonkar003: why 2id sqr = ad sqr

adityasonkar003: and why if ci bisects angle c therefore ci is perpendicular to ad

Answered by SyedNomanShah

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