Maths-
General
Easy

Question

ABC is a triangle in which AB = AC. P is any point in its interior such
thatstraight angle ABP equals straight angle ACP. Prove that AP bisects straight angle BAC

hintHint:

find relation between angle B and angle C using AB = AC. Now ,
Find relation between ∠PBC and ∠PCB  Also relation between PB and PC.
Now prove straight triangle A B P and  straight triangle A C P are congruent so , corresponding angles are equal.

The correct answer is: Hence proved


    In straight capital delta ABC,
    AB = AC (given)

     ∠B = ∠C(angles opposite to equal sides are equal)

    here, ∠ ABP  =  ∠ ACP (given)
    From diagram we get,

    ∠B = ∠ ABP  +  ∠ PBC —-Eq1

    ∠C = ∠ ACP  +  ∠ PCB—-Eq2
    Subtracting Eq1-Eq2 we get,

    ∠B - ∠C = ∠ ABP  +  ∠ PBC - ∠ ACP - ∠ PCB

    Substitute ∠  ABP  =  ∠ ACP and ∠ B  =  ∠ C,

    We get ∠ PBC  =  ∠ PCB

    In straight capital delta  PBC∠ PBC = ∠PCB

    Then , PB  = PC (side opposite to equal angles are equal)

    We get PB  =  PC(side )

    ∠ABP  =  ∠ ACP (given)(Angle)

    AB =  AC (given)(side)

    By SAS rule, straight triangle A B P approximately equal to straight triangle A C P
    By congruence , ∠ BAP  =  ∠ CAP then , ∠ BAC is bisected by AP
    Hence proved

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