Q.141.Properties of triangles.

Question 141.

1)

If the median of triangle through A is perpendicular to AB , prove that tan A + 2 tan B = 0.

2)

In a triangle , {1- (r1/r2)} {1-(r1/r3)} = 2, prove that the triangle is right angled.

r1 , r2 , r3 are the radii of escribed circle (The circles which touches the side BC and the two sides AB and AC extended is called ESCRIBED CIRCLE opposite to angle A.Its radius is denoted by r1.Similarly, r2 & r3 are the radii of the escribed circles opposite to the angle B & C respectively.).

3)

If p1, p2, p3 are the altitudes of a triangle from the vertices A, B, and C respectively and Δ represents the area of the triangle , prove that 1/p1 +1/p2 - 1/p3 = [2ab / {(a+b+c) Δ}] * cos ² (C/2).

Answer 141.

1)

Let AD be the median through A
=> AB = BDcosB
Applying sine rule to ΔABC,
AB/sinC = BC/sinA
=> BDcosB/sinC = 2BD/sinA
=> 2sinC - sinAcosB = 0
=> 2sin(A+B) - sinAcosB = 0 ... [because C = π - (A+B)]
=> 2sinAcosB + 2cosAsinB - sinAcosB = 0
=> sinAcosB + 2cosAsinB = 0
Dividing by cosAcosB,
=> tanA + 2tanB = 0.

2)
(1 - r1/r2)(1 - r1/r3) = 2
=> [1 - (s-b)/(s-a)] * [1 - (s-c)/(s-a)] = 2
=> (b-a)(c-a) = 2(s-a)^2
=> (b-a)(c-a) = 2[(b+c-a)/2]^2
=> 2(b-a)(c-a) = (b+c-a)^2
=> 2(bc -ca - ab + a^2) = b^2 + c^2 +a^2 + 2bc - 2ca - 2ab
=> a^2 =b^2 + c^2
=> ΔABC is a right triangle with ∠A = π/2.

3)
ap1 = bp2 = cp3 = 2Δ
=> LHS
= 1/p1 + 1/p2 - 1/p3
= a/2Δ + b/2Δ - c/2Δ
= (1/2)(a + b - c)/Δ
= (s-c)/Δ
RHS
= [2ab/{(a+b+c)*Δ}] * cos^2 (C/2)
= (ab/sΔ) * [s(s-c)/ab]
= (s-c)/Δ

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