I WILL MAKE BRAINLISEST In the given figure, ABC = AED = 90° and BD = DE.

(i) Name the locus of the points which are equi-distant from AB and AC.

(ii) If ACB = 40° , what are the magnitudes of BÂD and CÂD ?

(i) ∠A=∠A (common angle for both triangles)

∠ACB=∠ADE [given]

Therefore, △ABC∼△AED

(ii) from (i) proved that, △ABC∼△AED

So, BC/DE=AB/AE=AC/AD

AD=AB−BD

=6–1=5

Consider, AB/AE=AC/AD

6/3=AC/5

AC=(6×5)/3

AC=30/3

AC=10cm.

Step-by-step explanation:

Step-by-step explanation:

As per given details we can state:

Because:

Therefore:

i) The locus of the points which are equidistant from AB and AC is the line AD as it is the angle bisector of ∠BAC.

The point equidistant from AB and AC is point D

ii) ΔABC is a right triangle, find angle measure of ∠BAC:

Find the magnitudes of BÂD and CÂD:

3+x

step-by-step explanation:

150.6

all i did was i multiplied 5.02 by 30