ABCD is a parallelogram in which E and F are the midpoints of side BC and AD respectively. Lines are drawn from A to E and C to F such that AE = CF. Prove that the ratio of the sid... ABCD is a parallelogram in which E and F are the midpoints of side BC and AD respectively. Lines are drawn from A to E and C to F such that AE = CF. Prove that the ratio of the sides BC and BA are 2:1. Read more

Steps to Solve

1. Draw the Parallelogram and Mark Midpoints

Draw parallelogram ABCD. Mark E as the midpoint of BC and F as the midpoint of AD. Draw lines AE and CF.

2. Properties of Parallelograms

In parallelogram ABCD, opposite sides are equal and parallel. Thus, $BC = AD$ and $BC \parallel AD$, $AB = CD$ and $AB \parallel CD$.

3. Using Midpoint Properties

Since E and F are midpoints, $BE = \frac{1}{2}BC$ and $AF = \frac{1}{2}AD$.

Since $BC = AD$, then $\frac{1}{2}BC = \frac{1}{2}AD$, hence, $BE = AF$.

4. Parallel and Equal Segments

Since $BC \parallel AD$, then $BE \parallel AF$. Also, $BE = AF$. Therefore, quadrilateral AECF is a parallelogram because it has one pair of sides that are both parallel and equal.

5. Given Condition

We are given that $AE = CF$.

6. Parallelogram to Rectangle

Since AECF is a parallelogram and its diagonals are equal ($AE = CF$), AECF must be a rectangle. A parallelogram with equal diagonals is a rectangle.

7. Right Angle Implication

Because AECF is a rectangle, $\angle AFE = 90^\circ$. Since $AD \parallel BC$, $\angle AFE = \angle CBE$ (corresponding angles), thus $\angle CBE = 90^\circ$.

8. Parallelogram to Rectangle Again

Since $\angle CBE = 90^\circ$, parallelogram ABCD has a right angle, implying it is a rectangle.

9. Using Right Triangle ABE

Consider right triangle ABE. We have $AE^2 = AB^2 + BE^2$ by the Pythagorean theorem.

10. Substituting Known Values

Since $BE = \frac{1}{2}BC$ and $AE = CF$, and because ABCD is a rectangle, $AD=BC$ and $AB = CD$, also $AE = \sqrt{AB^2 + BE^2}$. Since AECF is a rectangle $AE = \sqrt{AF^2 + FE^2}$, but $FE = CD$ and $AF=BE$ so you could say $AE = \sqrt{BE^2 + CD^2}$ and we know $AB=CD$ so $AE = \sqrt{BE^2 + AB^2}$.

From the problem statement $AE = CF$

Since AECF is a rectangle $AE = CF$, so $AE = AE$. Then $AE^2 = AB^2 + BE^2$ $AE^2 = AB^2 + (\frac{1}{2}BC)^2$

Since ABCD is a rectangle $BC = AD$ and $AD = BC$, and also $AB = CD$

11. Relating BC and BA

Then $AE = CF$, so we can relate the sides. Since $AE = CF$, $AE^2 = CF^2$. Also in a rectangle $AE^2 = AB^2 + BE^2$ and $CF^2 = CD^2 + DF^2$. Since $AB = CD$ and $BE = DF$ we can confirm $AE = CF$. $AE^2 = AB^2 + (\frac{1}{2}BC)^2$, and $AE = CF$.

12. Final Equation

$AB^2+(\frac{1}{2}BC)^2 = AE^2$.

Since this is a rectangle we can relate the sides via $AE = CF$, $AE^2 = CF^2$ and since it makes a right angle $AE^2 = BE^2 + AB^2$, so $AE^2 = (\frac{1}{2}BC)^2 + AB^2$.

Given $AE=CF$, and $BE = \frac{1}{2}BC$, we solve:

if $BC = AD$, then $BE = AF$, and $AE = CF$ so $AE = CF = \sqrt{AB^2 + (\frac{1}{2}BC)^2}$. Because $AE = CF$, we can make the relationship to relate the sides of the rectangle.

$(\frac{1}{2}BC)^2 + AB^2 = CF^2=AE^2$. And since $\angle ABC = 90$ then $AE^2 = AB^2 + BE^2$, but $BE=\frac{1}{2}BC$ so $\frac{1}{4}BC^2+AB^2 = BC^2$, so $BC = 2AB$.

Then in $BC = 2BA$, $\frac{BC}{BA} = 2$.