Find area of a triangle inside of a parallelogram.

[ROCCOANDROXY]

Let m(angle(BAE)) = lambda and m(angle(EDC)) = m(angle(ECD)) = theta.

Triangle(AED) is an isosceles right triangle implies m(angle(EAD)) = m(angle(EDA)) = 45(degrees) and angle(A) and angle(D) are supplementary implies 45(degrees) + lambda + 45(degrees) + theta = 180(degrees) implies lambda = 90(degrees) - theta and isosceles Triangle(DEC) implies 5 = ED * cos(theta) implies ED = 5 * sec(theta) implies A(Triangle(ABE)) = 1/2 * (10) * (5 * sec(theta)) * cos(theta) = 25.

[Okkk]

A quick and easy way to find the answer ( without the beautiful insight presented here in this lesson ) is to consider the special case as the parallelogram becomes a rectangle (square). It can be shown that angle <DEC=90٥ and hence, AEC is a straight line as well as BED. Now clearly the green triangle area is a quarter of that of the square ABCD or 100/4=25.

[leonardoorellano]

Hello Science Sergei, first of all let me say that you have a great channel and i always look forward for a new video. The statement BAE + ADC =180 seems to be a typo. It should be BAD + ADC = 180 right ?

[panPetrff]

I did it the same way - passing perpendicular line to AB, CD through E. There are complementary angles at point E: (90°-α) + 90° + α = 180°
There is another problem to solve, a harder one. How can we find alpha and other angles? (or to prove that the triangle EDC is equilateral)