In $Δ$ ABC, $∠$ B $= 45^{o}$ , $∠$ C $= 65^{o}$ and the bisector of $∠$ BAC meets BC at P. Then the ascending order of sides is ___________.

AP, BP, CP

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AP, CP, BP

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BP, AP, CP

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None of these

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Solution

The correct option is D None of these
$∠ B A P = ∠ C A P = 35^{o}$ (because AP is angle bisector)
$∠ B > ∠ B A P \Rightarrow A P > B P$
and $∠ C > ∠ C A P \Rightarrow A P > C P$
$∠ A B C < ∠ A C P \Rightarrow C P < B P$
therefore ascending order is $C P, B P, A P$

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Similar questions

In triangle ABC, angle B = 35^o, angle C = 65^o and the bisector of angle BAC meets BC in P. Choose the correct descending order of AP, BP and CP

Q. In Δ ABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in P. Arrange AP, BP and CP in descending order.

BO and CO are respectively the bisectors of angle B and C of ∆ABC. AO produced meets BC at P. Show that AO/BP = AO/OP , AC/CP = AO/OP ,

AB/AC = BP/PC , AP is the bisector of angle BAC.

Q. In given figure BO and CO are respectively the bisectors of

∠ B

and

∠ C

△ A B C

. AO produced meets BC at P show that

(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) AP is the bisector of

∠ B A C

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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In ΔABC, ∠B=45o, ∠C=65o and the bisector of ∠BAC meets BC at P. Then the ascending order of sides is ___________.

The correct option is D None of these∠BAP=∠CAP=35o (because AP is angle bisector)∠B>∠BAP⇒AP>BPand ∠C>∠CAP⇒AP>CP∠ABC<∠ACP⇒CP<BPtherefore ascending order is CP,BP,AP

In $Δ$ ABC, $∠$ B $= 45^{o}$ , $∠$ C $= 65^{o}$ and the bisector of $∠$ BAC meets BC at P. Then the ascending order of sides is ___________.

The correct option is D None of these
$∠ B A P = ∠ C A P = 35^{o}$ (because AP is angle bisector)
$∠ B > ∠ B A P \Rightarrow A P > B P$
and $∠ C > ∠ C A P \Rightarrow A P > C P$
$∠ A B C < ∠ A C P \Rightarrow C P < B P$
therefore ascending order is $C P, B P, A P$