In Δ A B C , A C > A B .. The internal angle...

chivistaelmore

chivistaelmore

Answered

2022-07-23

In Δ A B C , A C > A B .. The internal angle bisector of A meets BC at D, and E is the foot of the perpendicular from B onto AD. Suppose A B = 5 , B E = 4 and A E = 3. Find ( A C + A B A C A B ) E D.

Answer & Explanation

nuramaaji2000fh

nuramaaji2000fh

Expert

2022-07-24Added 18 answers

Step 1
I use a bit of trigonometry.(I would love to see something without this)
Let B C = a , C A = b , A B = c.
By Napiers Analogy:
(1) tan B C 2 = b c b + c cot ( A / 2 )
Notice that
sin ( A / 2 ) = 4 5
(2) cot ( A / 2 ) = ?
Step 2
Also E D B = 90 + C B 2
(3) tan E D B = tan ( 90 + C B 2 ) = cot B C 2 = 4 E D
Jaxon Hamilton

Jaxon Hamilton

Expert

2022-07-25Added 3 answers

Step 1
This proof can probably be simplified or made more natural, but here is my take:

Construct CG parallel to FB as shown above. We have:
A F E A B E   ( A S A ) , A F E A C G , B E D C G D   ( A A A )
The last ratio is by Angle Bisector Theorem.
Step 2
Hence: A C + A B A C A B = A C C F + A B C F = A B C F ( 1 + A C A B ) = A F C F ( 1 + D G E D ) = A E E G E G E D = A E E D
Giving the result: ( A C + A B A C A B ) E D = A E

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