alan655
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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 WILL MAKE BRAINLISESTIn The Given Figure ABC AED 90 And BD DE I Name The Locus Of The Points Which Are Equidistant From AB And AC Ii If ACB 40 What Are The Ma class=

Sagot :

tryispy019

Answer:

(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:

mhanifa

Step-by-step explanation:

As per given details we can state:

  • ΔABD ≅ ΔAED

Because:

  • AD is common hypotenuse and BD = DE

Therefore:

  • ∠BAD ≅ ∠EAD or ∠BAD ≅ ∠CAD

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:

  • m∠BAC = 90° - 40° = 50°

Find the magnitudes of BÂD and CÂD:

  • m∠BAD = m∠CAD = (1/2)m∠BAC = (1/2)(50°) = 25°
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