Question

In the given figure the bisectors of angle B and angle C of a triangle ABC meet at I. IF IP⊥BC, IQ⊥CA and IR⊥AB, prove that (i) IP = IQ, (ii) IA bisects ∠A​

Answer 1

$\bf\underline{\underline{\pink{Question:-}}}$

★ the given figure the bisectors of angle B and angle C of a triangle ABC meet at I. IF IP⊥BC, IQ⊥CA and IR⊥AB, prove that (i) IP = IQ, (ii) IA bisects ∠A

$\bf\underline{\underline{\blue{Given:-}}}$

★In ∆ABC in which BI and CI are bisectors of ∠B and ∠C respectively. IP⊥BC, IQ⊥CS and IR⊥AB

$\bf\underline{\underline{\red{ToProve:-}}}$

★ IP = IQ = IR

★ IA bisects ∠A

$\bf\underline{\underline{\green{Proof:-}}}$

★IP = IQ = IR

In ∆IPC and ∆IQC we have,

∠IPC = ∠IQC = 90° (given)

∠ICP = ∠ICQ [ ∵ CI is the bisector of ∠C]

CI = CI (common)

∴ ∆IPC ≅ ∆IQC (A.S.A-criteria)

∴ IP = IR (c.p.c.t)

Hence, IP = IQ = IR.

★IA bisects ∠A

In ∆IQA and ∆IRA, we have

IQ = IR [proved in (i)]

∠IQA = ∠IRA [each = 90°]

hyp. IA = hyp. IS (common)

∴ ∆IQA ≅ ∆IRA (R.H.S-criteria)

∴ ∠IAQ = ∠IAR (c.p.c.t).

Hence, iA bisects∠A.

Answer 2

Given :- In the given figure the bisectors of angle B and angle C of a triangle ABC meet at I. IP⊥BC, IQ⊥CA and IR⊥AB .

To Prove :-

• IP = IQ .
• IA bisects ∠A .

Solution :-

(i) Proof :-

In ∆IPC and ∆IQC we have,

→ ∠ICP = ∠ICQ { CI is angle bisector of ∠C .}

→ ∠IPC = ∠IQC { given that, IP⊥BC, IQ⊥CA . So, both are equal to 90°. }

→ IC = IC { Common }

So,

→ ∆IPC ≅ ∆IQC { By AAS congruence rule . }

then,

→ IP = IQ { When two ∆'s are congruent their corresponding sides are equal in length . }

therefore,

→ IP = IQ {Proved.}

Similarly, we can prove that,

→ IP = IR

hence, we can conclude that,

→ IP = IR = IQ --------- Equation (1)

(ii) Proof :-

In ∆AIR and ∆AIQ we have,

→ ∠ARI = ∠AQI { Since IQ⊥CA and IR⊥AB , both are equal to 90° .}

→ AI = AI { common , side opposite to right angle ∆ is known as hypotenuse . }

→ IR = IQ { from Equation (1) }

So,

→ ∆AIR ≅ ∆AIQ { Two ∆'s are congruent by RHS congruence rule when lengths of hypotenuse and one side of both ∆'s are equal . }

then,

→ ∠IAR = ∠IAQ { When two ∆'s are congruent their corresponding sides are equal in length . }

therefore,

→ IA bisects ∠A {Proved.}

Learn more :-

In the figure along side, BP and CP are the angular bisectors of the exterior angles BCD and CBE of triangle ABC. Prove ∠BOC = 90° - (1/2)∠A .

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