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Problem 201

In the figure, \(\mathrm{ABCD}\) is a parallelogram with diagonals \(\underline{\mathrm{AC}}\) and BD. \(\angle \mathrm{ABC}\) is an obtuse angle. Prove that \(\mathrm{AC}>\mathrm{BD}\).

Expert verified

In the parallelogram ABCD with an obtuse angle at vertex B, we can prove that \(AC > BD\) using the triangle inequality theorem applied to triangles ABC and ADC. Since AC is the longest side in both triangles due to the obtuse angles, we find that \(BD + AC > 2 \cdot AC\), which can be rearranged as \(BD > AC\).

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Chapter 12

Let \(\mathrm{ABCD}\) be a rhombus. Prove that diagonal \(\underline{\mathrm{AC}}\) bisects \(\angle \mathrm{A}\).

Chapter 12

Prove that both pairs of opposite sides of a parallelogram are congruent.

Chapter 12

Prove that a rectangle is a parallelogram.

Chapter 12

Starting with any triangle \(\mathrm{ABC}\); construct the exterior squares \(\mathrm{BCDE}, \mathrm{ACFG}\) and BAHK; then construct parallelograms \(\mathrm{FCDQ}\) and EBKP. Prove \(\triangle \mathrm{PAQ}\) is an isosceles right triangle. (Hint: Draw diagonals \(\mathrm{PB}\) and CQ.)

Chapter 12

Given: Square \(\mathrm{ABCD} ; \mathrm{P}\) is any point of \(\underline{\mathrm{AB}}, \mathrm{Q}\) is any point $\underline{\mathrm{AD}}, \underline{\mathrm{CQ}} \perp \underline{\mathrm{PD}}\( at \)\mathrm{R}$. Prove \(\underline{\mathrm{PD}} \cong \mathrm{QC}\).

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