Question

Let ABC be a triangle with angle A = 90 and AB = AC. Let D and E be points on the segment
BC such that BD : DE : EC = 3 : 5 : 4. Prove that angle DAE = 45.

Answer 1

Answer on Question #54217, Math / Geometry

Given:

Let ABC be a triangle with angle A = 90 and AB = AC. Let D and E be points on the segment

BC such that BD : DE : EC = 3 : 5 : 4. Prove that angle DAE = 45 .

Solution:

BD : DE : EC = 3 : 5 : 4 ⇒ ∠BAD : ∠DAE : ∠EAC = 9 : 25 : 16

So we obtain

$$9x^2 + 25x^2 + 16x^2 = 90 \quad ⇒ \quad 50x^2 = 90$$ $$x^2 = \frac{90}{50} = 1.8$$ $$\angle DAE = 25x^2 = 25 \cdot 1.8 = 45{}^\circ$$

Answer: ∠DAE = 45°

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