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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 \(\mathrm{AC}>\mathrm{BD}\).

Short Answer

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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Step 1: 1. Properties of a parallelogram

Since ABCD is a parallelogram, opposite sides are equal. Therefore, AB = CD, and BC = AD. Also, opposite angles are equal, which means that if angle ABC is obtuse, angle CDA is also obtuse.

Step 2: 2. Triangles ABC and ADC

By drawing the diagonals AC and BD in parallelogram ABCD, we also create two triangles: ABC and ADC. Since AC is a diagonal, it is a common side for both triangles.

Step 3: 3. Using the Triangle Inequality theorem

According to the Triangle Inequality theorem, in any triangle, the sum of the lengths of any two sides must be greater than the length of the remaining side. Applying this theorem to triangles ABC and ADC, we have: In triangle ABC: \(AB + BC > AC\) (1) In triangle ADC: \(AD + DC > AC\) (2)

Step 4: 4. Obtuse angle and longer side

In any triangle, the side opposite an obtuse angle is always the longest side. Since angle ABC and angle CDA are both obtuse, it means that side AC is the longest side in both triangles ABC and ADC.

Step 5: 5. Replacing sides with equal sides from the parallelogram

Since AB = CD and BC = AD (from step 1), we can replace the sides in equations (1) and (2) with their equal sides: In triangle ABC: \(CD + AD > AC\) (1) In triangle ADC: \(BC + AB > AC\) (2)

Step 6: 6. Adding the inequalities

Adding the inequalities (1) and (2), we get: \((CD + AD) + (BC + AB) > 2 \cdot AC\)

Step 7: 7. Combining the terms and simplifying

Now, we can combine the terms: \((CD + AD) + (BC + AB) = (AB + BC) + (AD + CD) = (AB + AD) + (BC + CD)\) And since AB + AD = BD and BC + CD = AC, the inequality becomes: \((AB+AD)+(BC+CD)=BD+AC>2\cdot AC\)

Step 8: 8. Proving AC > BD

Rearranging the last inequality to isolate AC on one side: \(BD+AC>2\cdot AC\) Now, subtract AC from both sides: \(BD > AC\) Therefore, it is proved that AC is longer than BD in the given parallelogram with obtuse angles at vertices B and D.

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