BO and CO are respectively the bisectors of - B and - C of - ABC- AO produced meets BC at P-Show that -4 MARKSi AB - BP - AO - OPii AC - CP - AO - OPiii AB - AC - BP - PCiv AP is the bisector of - B A C-

Each proof : 1 Mark each (i) In Δ ABP, BO is the bisector of ∠ B ∴AB/BP = AO/OP [Angle bisector theorem] (ii) In Δ ACP, OC is the bisector of ∠ C. ∴AC/CP ...

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Byju's Answer

Standard X

Mathematics

Perpendicular from Right Angle to Hypotenuse Divides the Triangle into Two Similar Triangles

BO and CO are...

Question

BO and CO are respectively the bisectors of $∠ B$ and $∠ C$ of $Δ A B C$ . AO produced meets BC at P.

Show that [4 MARKS]

(i) $\frac{A B}{B P} = \frac{A O}{O P}$

(ii) $\frac{A C}{C P} = \frac{A O}{O P}$

(iii) $\frac{A B}{A C} = \frac{B P}{P C}$

(iv) $A P$ is the bisector of $∠ B A C$ .

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Solution

Each proof : 1 Mark each

(i) In $Δ A B P$ , BO is the bisector of $∠ B$

$∴ \frac{A B}{B P} = \frac{A O}{O P}$ [Angle bisector theorem]

(ii) In $Δ A C P$ , OC is the bisector of $∠ C$ .

$∴ \frac{A C}{C P} = \frac{A O}{O P}$ [Angle bisector theorem]

(iii) We have, proved that

$\frac{A B}{B P} = \frac{A O}{O P}$ and $\frac{A C}{C P} = \frac{A O}{O P}$

$\Rightarrow \frac{A B}{B P} = \frac{A C}{C P}$

$\Rightarrow \frac{A B}{A C} = \frac{B P}{P C}$

(iv) As proved above that in $Δ A B C$ , we have

$\frac{A B}{A C} = \frac{B P}{P C}$

$\Rightarrow A P$ is the bisector of $∠ B A C$ .

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[4 MARKS]

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BO and CO are respectively the bisectors of ∠B and ∠C of ΔABC. AO produced meets BC at P. Show that [4 MARKS] (i) ABBP=AOOP (ii) ACCP=AOOP (iii) ABAC=BPPC (iv) AP is the bisector of ∠BAC.

BO and CO are respectively the bisectors of $∠ B$ and $∠ C$ of $Δ A B C$ . AO produced meets BC at P.

Show that [4 MARKS]

(i) $\frac{A B}{B P} = \frac{A O}{O P}$

(ii) $\frac{A C}{C P} = \frac{A O}{O P}$

(iii) $\frac{A B}{A C} = \frac{B P}{P C}$

(iv) $A P$ is the bisector of $∠ B A C$ .