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Sia 🤖 1 week, 3 days ago

In {tex}\triangle{/tex}ABC,

BP bisects {tex}\angle{/tex}ABC

{tex}\therefore{/tex} {tex}\angle{/tex}ABP = {tex}\angle{/tex}PBC = {tex}\frac{1}{2}{/tex}{tex}\angle{/tex}B ={tex}\frac{1}{2}{/tex}(2{tex}\angle{/tex}C) = {tex}\angle{/tex}C . . . . (1)

In PBC,

{tex}\therefore{/tex} {tex}\angle{/tex}PBC = {tex}\angle{/tex}PCB (= ∠C)

{tex}\therefore{/tex} PB = PC . . . [Sides opposite to equal angles] . . .(2)

In {tex}\triangle{/tex}APB and {tex}\triangle{/tex}DPC,

AB = CD . . . [Given]

PB = PC . . .[From (2)]

{tex}\angle{/tex}ABP = {tex}\angle{/tex}DCP (= {tex}\angle{/tex}C)

{tex}\therefore{/tex} {tex}\triangle{/tex}APB = {tex}\triangle{/tex}DPC . . . . [By SAS property]

{tex}\therefore{/tex} {tex}\angle{/tex}BAP = CDP (= {tex}\angle{/tex}A) . . .[c.p.c.t.] . . . (3)

and AP = DP . . . [c.p.c.t.] . . . .(4)

In {tex}\triangle{/tex}APD,

As AP = DC . . . [From (4)]

{tex}\therefore{/tex} {tex}\angle{/tex}PDA = {tex}\angle{/tex}PAD = {tex}\frac{A}{2}{/tex}

{tex}\therefore{/tex} {tex}\angle{/tex}DPA = {tex}\pi-\left(\frac{A}{2}+\frac{A}{2}\right)=\pi-A{/tex} . . . . (5)

From {tex}\angle{/tex}DPC

{tex}\angle{/tex}DPC = {tex}\pi{/tex} – (A + C)

{tex}\therefore{/tex} {tex}\angle{/tex}DPA = {tex}\pi{/tex} – {tex}\angle{/tex}DPC = {tex}\pi{/tex} – [{tex}\pi{/tex} – (A + C)] = A + C . . .(6)

From (5) and (6)

{tex}\pi{/tex} – A = A + C {tex}\therefore{/tex} 2A + C = {tex}\pi{/tex} . . . (7)

Again

A + B + C = {tex}\pi{/tex} . . .[Sum of three angles of a triangle = {tex}\pi{/tex}]

{tex}\therefore{/tex} A + 2C + C = {tex}\pi{/tex} . . .[As B = 2C]

{tex}\therefore{/tex} A + 3C = {tex}\pi{/tex} . . . (8)

Multiplying (7) by 3, we get

6A + 3C = 3{tex}\pi{/tex}

5A = 2{tex}\pi{/tex} . . .[By subtracting (8) from (9)]

{tex}\therefore{/tex} {tex}A=\frac{2 \pi}{5}=\frac{2}{5} \times 180^{\circ}=72^{\circ}{/tex}

{tex}\therefore{/tex} {tex}\angle{/tex}BAC = 72

^{o}0Thank You