alan655
Answered

Welcome to Westonci.ca, your go-to destination for finding answers to all your questions. Join our expert community today! Join our platform to connect with experts ready to provide detailed answers to your questions in various areas. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

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°
We hope our answers were helpful. Return anytime for more information and answers to any other questions you may have. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.