HELP! I don t understand how to prove this math problem.?

Given ∆ABC with AB>AC. The bisectors of the interior and exterior angles at A intersect BC at points D and E, respectively.

Prove that (DE/CD)-(DE/BD)=2

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  • atsuo
    Lv 6
    3 years ago
    Favorite Answer

    Please draw a diagram .

    ................. A

    ............. * * * ..... *

    ....... * ... * . * .............. *

    B ------ D -- C ------------------ E

    AD is a bisector of ∠BAC , so ∠BAD = ∠CAD = (1/2)∠BAC .

    AE is a bisector of the exterior angle of ∠BAC , so

    ∠CAE = (1/2)(180°- ∠BAC)

    = (1/2)(180°- 2*∠BAD)

    = 90° - ∠BAD

    = 90° - θ (represent ∠BAD as θ . ∠CAD = θ , too .)

    Next , we know the angle bisector theorem , it says BD : CD = AB : AC .

    Let AC = 1 and AB = k (k > 1) , so BD : CD = k : 1 , therefore BD = kCD .

    Think △ABD . By the cosine law ,

    BD^2 = AB^2 + AD^2 - 2AB*ADcosθ

    (kCD)^2 = k^2 + AD^2 - 2k*ADcosθ

    k^2*CD^2 = k^2 + AD^2 - 2k*ADcosθ ---(#1)

    Think △ACD . By the cosine law ,

    CD^2 = AC^2 + AD^2 - 2AC*ADcosθ

    CD^2 = 1^2 + AD^2 - 2ADcosθ ---(#2)

    From (#2) ,

    2ADcosθ = 1 + AD^2 - CD^2 ---(#3)

    Substitute (#3) into (#1) ,

    k^2*CD^2 = k^2 + AD^2 - k(1 + AD^2 - CD^2)

    k^2*CD^2 = k^2 + AD^2 - k - kAD^2 + kCD^2

    kAD^2 - AD^2 = k^2 - k - k^2CD^2 + kCD^2

    (k - 1)AD^2 = k(k - 1) - CD^2*k(k - 1)

    ... k > 1 so k - 1 ≠ 0 ,

    AD^2 = k - kCD^2

    AD^2 = k(1 - CD^2)

    AD = √(k(1 - CD^2)) ---(#4)

    Substitute (#4) into (#3) ,

    2√(k(1 - CD^2))cosθ = 1 + (k(1 - CD^2)) - CD^2

    2√(k(1 - CD^2))cosθ = (1 + k) - (1 + k)CD^2

    2√(k(1 - CD^2))cosθ = (1 + k)(1 - CD^2)

    cosθ = (1 + k)(1 - CD^2) / [2√(k(1 - CD^2))] ---(#5)

    Think △ACE . By the sine law ,

    CE / sin(∠CAE) = AC / sin(∠AEC)

    CE / sin(90° - θ) = 1 / sin(∠AEC)

    CEsin(∠AEC) = sin(90° - θ)

    CEsin(∠AEC) = cosθ

    sin(∠AEC) = cosθ / CE ---(#6)

    Think △ADE , it is a right triangle ,

    AD = DEsin(∠AEC) ---(#7)

    Substitute (#4) and (#6) into (#7) ,

    √(k(1 - CD^2)) = DEcosθ / CE

    CE√(k(1 - CD^2)) / DE = cosθ ---(#8)

    By (#5) and (#8) ,

    CE√(k(1 - CD^2)) / DE = (1 + k)(1 - CD^2) / [2√(k(1 - CD^2))]

    CE√(k(1 - CD^2))*[2√(k(1 - CD^2))] = DE(1 + k)(1 - CD^2)

    2CE(k(1 - CD^2)) = DE(1 + k)(1 - CD^2)

    ... (1 - CD^2) ≠ 0 by (#4) ,

    2kCE = DE(1 + k)

    ... CE = DE - CD , so

    2k(DE - CD) = DE(1 + k)

    2kDE - 2kCD = DE(1 + k)

    DE(2k - (1 + k)) = 2kCD

    DE(k - 1) = 2kCD

    DE = 2kCD / (k - 1) ---(#9)

    By (#9) ,

    DE / CD - DE / BD

    = (2kCD / (k - 1)) / CD - (2kCD / (k - 1)) / kCD (because BD = kCD)

    = 2k / (k - 1) - 2 / (k -1)

    = 2(k - 1) / (k - 1)

    = 2

    So it is proved .

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