Question

Did I do this right?

Answer 1

Yes I think so. AD¯¯¯¯¯ , BD¯¯¯¯¯ , and CD¯¯¯¯¯ are angle bisectors of the vertex angles of △ABC . CF=8 meters and CD=17 meters.
What is DE ? The answer is 15m.

Answer 2

Answer : The value of side DE is, 15 m

Step-by-step explanation :

First we have to determine the side DF.

Using Pythagoras theorem in ΔDFC :

$(Hypotenuse)^2=(Perpendicular)^2+(Base)^2$

$(CD)^2=(DF)^2+(CF)^2$

Given:

Side CD = 8 m

Side CF = 17 m

Now put all the values in the above expression, we get the value of side DF.

$(8)^2=(DF)^2+(17)^2$

$DF=\sqrt{(17)^2-(8)^2}$

$DF=15m$

Now we have to determine the value of side DE.

Side DE = Side DF = 15 m   (By congruent triangle DFB and DEB)

The ΔDFB and ΔDEB are congruent triangles.

Proof:

Side DB = Side DB  (common side)

∠DBE = ∠DBF    (BD is angle bisector, so these angle are always equal)

∠E = ∠F    (equal to 90°)

ΔDFB ≅ ΔDEB    (By SAS congruency)

So, Side DE = Side DF

Hence, the value of side DE is, 15 m