The sides AB and AC of a triangle ABC are produced; and the bisectors of the external angles at B and C meet at P Now if AB >
AC, then PC < PB.
If the above statement is true then mention answer as 1, else mention 0 if false

Question

The sides AB and AC of a triangle ABC are produced- and the bisectors of the external angles B and C meet P- Prove that AB - AC- then PC - PB-

Toppr Admin · Accepted Answer

Here AB-gt-AC-lt-ACB-gt-lt-ABCAngle opposite to greater side is greater-lt-ECB-lt-lt-DBC -linear angle-12-lt-ECB-lt-12-lt-DBC-lt-PCB-lt-lt-PBC-x2235-PB-lt-PCPC-gt-PB Hence proved-

The sides $A B$ and $A C$ of a triangle $A B C$ are produced; and the bisectors of the external angles at $B$ and $C$ meet at $P$ . Prove that if $A B > A C$ , then $P C > P B$ .

Here AB $>$ AC
$< A C B >< A B C$
Angle opposite to greater side is greater
$< E C B << D B C$ (linear angle)
$\frac{1}{2} < E C B < \frac{1}{2} < D B C$
$< P C B << P B C$
$∵ P B < P C$
$P C > P B$ Hence proved.

The sides AB and AC of a triangle ABC are produced; and the bisectors of the external angles at B and C meet at P. Prove that if AB>AC, then PC>PB.

Here AB>AC<ACB><ABCAngle opposite to greater side is greater<ECB<<DBC (linear angle)12<ECB<12<DBC<PCB<<PBC∵PB<PCPC>PB Hence proved.

The sides $A B$ and $A C$ of a triangle $A B C$ are produced; and the bisectors of the external angles at $B$ and $C$ meet at $P$ . Prove that if $A B > A C$ , then $P C > P B$ .

Here AB $>$ AC
$< A C B >< A B C$
Angle opposite to greater side is greater
$< E C B << D B C$ (linear angle)
$\frac{1}{2} < E C B < \frac{1}{2} < D B C$
$< P C B << P B C$
$∵ P B < P C$
$P C > P B$ Hence proved.