Explanation:
The altitude CH divides triangle ABC into similar triangles:
ΔABC ~ ΔACH ~ ΔCBH
Angle bisector AL divides the triangle(s) into proportional parts:
BL/BA = CL/CA
HD/HA = CD/CA
Of course, the Pythagorean theorem applies to the sides of each right triangle:
AH^2 +CH^2 = AC^2
DH^2 +AH^2 = AD^2
LC^2 + AC^2 = LA^2
AC^2 +BC^2 = AB^2
And segment lengths sum:
HD +DC = HC
AD +DL = AL
AH +HB = AB
CL +LB = CB
Solving the problem involves picking the relations that let you find something you don't know from the things you do know. You keep going this way until the whole geometry is solved (or, at least, the parts you care about).
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We can use the Pythagorean theorem to find AH right away, since we already know AD and DH.
DH^2 +AH^2 = AD^2
4^2 + AH^2 = 8^2 . . . . . . . substitute known values
AH^2 = 64 -16 = 48 . . . . . . subtract 16
AH = 4√3 . . . . . . . . . . . . . . take the square root
Now, we can use this with the angle bisector relation to tell us how CD and CA are related.
HD/HA = CD/CA
4/(4√3) = CD/CA . . . . . substitute known values
CA = CD·√3 . . . . . . . . . cross multiply and simplify
Using the sum of lengths equation, we have ...
CH = HD +CD
CH = 4 + CD
From the Pythagorean theorem ...
AH^2 +CH^2 = AC^2
(4√3)^2 + (4 +CD)^2 = (CD√3)^2 . . . . . substitute known values
48 + (16 +8·CD +CD^2) = 3·CD^2 . . . . . simplify a bit
2·CD^2 -8·CD -64 = 0 . . . . . . . . . . . . . . . put the quadratic into standard form
2(CD -8)(CD +4) = 0 . . . . . . . . . . . . . . . . factor
CD = 8 . . . . . only the positive solution is useful here
Now, we know ...
∆ADC is isosceles, so ∠ACH = ∠CAD = ∠DAH = ∠CBA
CH = 8+4 = 12
AC = 8√3 . . . . . = 2·AH
Then by similar triangles, ...
AB = 2·AC = 16√3
BC = AC·√3 = 24
