R, P, Q act on the base AB, BC, CA triangles ABC in alphabetical order, respectively. Show that P sec A + Q sec B + R sec C = 0, if the fluid of this power system passes through the center of the triangle.

(i) In △ABD and △ACD,

AB=AC (since △ABC is isosceles)

AD=AD (common side)

BD=DC (since △BDC is isosceles)

ΔABD≅ΔACD SSS test of congruence,

∴∠BAD=∠CAD i.e. ∠BAP=∠PAC [c.a.c.t](i)

(ii) In △ABP and △ACP,

AB=AC ...(since △ABC is isosceles)

AP=AP ...(common side)

∠BAP=∠PAC from (i)

△ABP≅△ACP SAS test of congruence

∴BP=PC ...[c.s.c.t](ii)

∠APB=∠APC c.a.c.t.

(iii) Since △ABD≅△ACD

∠BAD=∠CAD from (i)

So, AD bisects ∠A

i.e. AP bisects∠A(iii)

In △BDP and △CDP,

DP=DP ...common side

BP=PC ...from (ii)

BD=CD ...(since △BDC is isosceles)

△BDP≅△CDP SSS test of congruence

∴∠BDP=∠CDP c.a.c.t.

∴ DP bisects∠D

So, AP bisects ∠D (iv)

From (iii) and (iv),

AP bisects ∠A as well as ∠D.

(iv) We know that

∠APB+∠APC=180

∘

(angles in linear pair)

Also, ∠APB=∠APC ...from (ii)

∴∠APB=∠APC=

2

180

∘

=90

∘

BP=PC and ∠APB=∠APC=90

∘

Hence, AP is perpendicular bisector of

Step-by-step explanation:

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