QUESTION IMAGE
Question
diagonals $overline{ac}$ and $overline{bd}$ form right angles at point m in parallelogram abcd. prove abcd is a rhombus.
statements
- abcd is a parallelogram
- $angle amb$, $angle bmc$, $angle cmd$, and $angle dma$ are right angles
- $angle ambcongangle bmccongangle cmdcongangle dma$
- ac bisects bd;
bd bisects ac;
- $overline{am}congoverline{mc}$, $overline{mb}congoverline{md}$
6.
reasons
- given
- given
- right angles are congruent
- diagonals of a parallelogram bisect each other
- definition of a bisector
- sas congruency theorem
Step1: Recall properties of parallelogram
Given that \(ABCD\) is a parallelogram, its diagonals \(AC\) and \(BD\) bisect each other at \(M\) (diagonals of a parallelogram bisect each other), so \(AM = MC\) and \(MB=MD\). Also, \(\angle AMB=\angle BMC = \angle CMD=\angle DMA = 90^{\circ}\) (given).
Step2: Consider triangles \(\triangle AMB\) and \(\triangle BMC\)
In \(\triangle AMB\) and \(\triangle BMC\), we have \(AM = MC\) (from diagonal - bisecting property), \(\angle AMB=\angle BMC = 90^{\circ}\), and \(MB\) is common to both triangles.
Step3: Apply SAS congruency theorem
By the Side - Angle - Side (SAS) congruency theorem (\(AM = MC\), \(\angle AMB=\angle BMC\), \(MB = MB\)), \(\triangle AMB\cong\triangle BMC\).
Step4: Use congruent - triangle property
Since \(\triangle AMB\cong\triangle BMC\), then \(AB = BC\) (corresponding parts of congruent triangles are congruent).
Step5: Recall the definition of a rhombus
A parallelogram with adjacent sides equal is a rhombus. Since \(ABCD\) is a parallelogram and \(AB = BC\), \(ABCD\) is a rhombus.
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In statement 6, the two triangles that can be used to apply the SAS congruency theorem are \(\triangle AMB\) and \(\triangle BMC\) (or \(\triangle BMC\) and \(\triangle CMD\), or \(\triangle CMD\) and \(\triangle DMA\), or \(\triangle DMA\) and \(\triangle AMB\)). Since \(\triangle AMB\cong\triangle BMC\), \(AB = BC\). A parallelogram with adjacent - sides equal is a rhombus, so \(ABCD\) is a rhombus.