In triangle ABC, the angle bisector of angle BAC meets line segment BC at D, such that AD = AB. Line segment AD is extended to E, such that CD = CE and angle DBE = angle BAD. Show that triangle ACE is isosceles. Please try to describe step-by-step how you achieved the answer, because i want to learn. :) Thank you for your help in advance!

DE = 11.25 units

Step-by-step explanation:

The image of the quadilateral BCED and the extensions of BD and CE to move past B and C to meet at A is presented in the attached file to this answer.

Let the angle at A be θ

Let the length of DE be x

ABC forms a triangle which van allow θ to be obtained using cosine rule.

BC² = AB² + AC² - (2×AB×AC×cos θ)

BC = 9

AB = 24

AC = 28

9² = 24² + 28² - (2×24×28×cos θ)

-1279 = -1344 cos θ

Cos θ = (1279/1344) = 0.9516

θ = 17.89°

Note that ADE also forms a triangle and we can also use cosine rule to obtain the required side DE

DE² = AD² + AE² - (2×AD×AE×cos θ)

DE = ?

AD = AB + BD = 24 + 11 = 35

AE = AC + CE = 28 + 2 = 30

θ = 17.89°

DE² = 35² + 30² - (2×35×30×cos 17.89°)

DE² = 126.54

DE = √126.54 = 11.25

Hope this Helps!!!

answer: i oofed

step-by-step explanation:

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