In a parallelogram ABCD, AB=10 cm, BC=14 cm, and angle B is 50 degrees. D and E are the midpoints of AB and AC respectively. Calculate DE and angle ADE.

In $\triangle ABC$, $D$ is the midpoint of $AB$ and $E$ is the midpoint of $AC$. By the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.

4.2.2

Since $D$ and $E$ are midpoints, $DE \parallel BC$ by the Midpoint Theorem. When two parallel lines are intersected by a transversal ($AB$), the corresponding angles are equal.

Reason: Corresponding angles ($DE \parallel BC$).

Question 5

5.1

In parallelogram $ABCD$, $AB \parallel DC$ and $AB = DC$ (opposite sides of a parallelogram are parallel and equal).

Given $BP \perp AC$ and $DQ \perp AC$, this means $\angle APB = 90^\circ$ and $\angle CQD = 90^\circ$.

Consider $\triangle ABP$ and $\triangle CDQ$:

• $\angle APB = \angle CQD = 90^\circ$ (Given $BP \perp AC$ and $DQ \perp AC$)
• $AB = CD$ (Opposite sides of parallelogram $ABCD$)
• $\angle BAP = \angle DCQ$ (Alternate interior angles, since $AB \parallel DC$ and $AC$ is a transversal) Therefore, $\triangle ABP \equiv \triangle CDQ$ (AAS congruence criterion).

5.2

Since $\triangle ABP \equiv \triangle CDQ$ (proven in 5.1), their corresponding sides are equal.
Therefore, $BP = DQ$.

Given $BP \perp AC$ and $DQ \perp AC$. If two lines are perpendicular to the same line, then they are parallel to each other.
Therefore, $BP \parallel DQ$.

Since one pair of opposite sides ($BP$ and $DQ$) in quadrilateral $BPCQ$ is both equal in length and parallel, $BPCQ$ is a parallelogram.