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

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

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

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

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

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

∠CAE = (1/2)(180°- ∠BAC)

= 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 ,

Think △ACD . By the cosine law ,

From (#2) ,

Substitute (#3) into (#1) ,

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

... k > 1 so k - 1 ≠ 0 ,

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 ,

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 .