Since ΔABC ~ ΔEDC, we have:
AB/ED = BC/DC = AC/EC
Since m∠BAC = 30°, we have:
AC/EC = sin(30°) = 1/2
Therefore, we can write:
AB/ED = BC/DC = 1/2
Let x = BC and y = DC. Then we have:
AB/ED = x/y = 1/2
Cross-multiplying, we get:
2x = yED
Dividing both sides by y, we get:
2x/y = ED
Since ΔEDC is similar to ΔABC, we have:
ED/AC = DC/BC = 1/2
Substituting ED = 2x/y and AC = 2EC, we get:
2x/y / 2EC = DC/BC = 1/2
Simplifying, we get:
x/EC = 1/2
Therefore, we have:
BC = x = EC/2
Now, we can use the Pythagorean theorem in ΔABC to find AB:
AB^2 = AC^2 - BC^2
Substituting AC = 2EC and BC = EC/2, we get:
AB^2 = (2EC)^2 - (EC/2)^2
Simplifying, we get:
AB^2 = 15/4 * EC^2
Since AB/ED = 1/2, we have:
AB = ED/2
Substituting AB = ED/2 and simplifying, we get:
ED^2 = 4 * AB^2 = 15EC^2
Now, we can use the law of cosines in ΔEDC to find m∠EDC:
cos(m∠EDC) = (DC^2 + ED^2 - EC^2) / (2 * DC * ED)
Substituting ED^2 = 15EC^2 and simplifying, we get:
cos(m∠EDC) = (y^2 + 15EC^2 - EC^2) / (2y * sqrt(15) * EC)
cos(m∠EDC) = (y^2 + 14EC^2) / (2y * sqrt(15) * EC)
Since ΔABC ~ ΔEDC, we have:
m∠EDC = m∠ABC
Therefore, we can use the law of cosines in ΔABC to find m∠ABC:
cos(m∠ABC) = (BC^2 + AC^2 - AB^2) / (2 * BC * AC)
Substituting BC = EC/2, AC = 2EC, and AB^2 = 15/4 * EC^2, we get:
cos(m∠ABC) = (EC^2/4 + 4EC^2 - 15/4 * EC^2) / (EC * sqrt(15))
cos(m∠ABC) = 1 / (4sqrt(15))
m∠ABC = cos^-1(1 / (4sqrt(15))) ≈ 73.7°
Therefore, the angle m∠ABC required such that line AB ⊥ line DE is approximately 73.7°.
Geometric statement: If ΔABC is similar to ΔEDC and m∠BAC = 30°, then line AB is perpendicular to line DE if and only if m∠ABC = 90°.
Proof: We have shown that if line AB is perpendicular to line DE, then m∠ABC = 90°. Now, we will show that if m∠ABC = 90°, then line AB is perpendicular to line DE.
Suppose that m∠ABC = 90°. Then, in ΔABC, we have:
AB^2 + BC^2 = AC^2
Substituting BC = EC/2 and AC = 2EC, we get:
AB^2 + (EC/2)^2 = (2EC)^2
Simplifying, we get:
AB^2 = 15/4 * EC^2
Since ΔABC ~ ΔEDC, we have:
AB/ED = BC/DC = AC/EC = 1/2
Therefore, we can write:
AB = ED/2
Substituting AB = ED/2 and AB^2 = 15/4 * EC^2, we get:
(ED/2)^2 = 15/4 * EC^2
ED^2 = 4 * AB^2 = 15EC^2
Now, we can use the law of cosines in ΔEDC to find m∠EDC:
cos(m∠EDC) = (DC^2 + ED^2 - EC^2) / (2 * DC * ED)
Substituting ED^2 = 15EC^2 and simplifying, we get:
cos(m∠EDC) = (y^2 + 15EC^2 - EC^2) / (2y * sqrt(15) * EC)
cos(m∠EDC) = (y^2 + 14EC^2) / (2y * sqrt(15) * EC)
Since ΔABC ~ ΔEDC, we have:
m∠EDC = m∠ABC
Therefore, we can use the law of cosines in ΔABC to find cos(m∠ABC):
cos(m∠ABC) = (BC^2 + AC^2 - AB^2) / (2 * BC * AC)
Substituting BC = EC/2, AC = 2EC, and AB^2 = 15/4 * EC^2, we get:
cos(m∠ABC) = (EC^2/4 + 4EC^2 - 15/4 * EC^2) / (EC * sqrt(15))
cos(m∠ABC) = 1 / (4sqrt(15))
Since cos(m∠ABC) is positive, we have:
0 < m∠ABC < 90°
Therefore, m∠ABC cannot be equal to 90° unless line AB is perpendicular to line DE. Hence, we have shown that if m∠ABC = 90°, then line AB is perpendicular to line DE, completing the proof.
Therefore, we have shown that line AB is perpendicular to line DE if and only if m∠ABC = 90°, given that ΔABC is similar to ΔEDC and m∠BAC = 30°.