Igot all of the answers but i can't explain it at explain guys i have no idea! problem: in right △abc, the altitude ch to the hypotenuse ab intersects angle bisector al in point d. find the sides of △abc if ad = 8 cm and dh = 4 cm. answers: ab = 16√3, ac = 8√3, bc =24

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