Question

1. (ls6) if a quadrilateral does not have four right angles, then it is not a rectangle.

a. inverse
b. contrapositive
c. converse
d. either a or c
e. none of the above

1. select the correct proof from the options listed.

given: rectangle (abcd); (m) is the midpoint of (overline{ad})
prove: (\triangle bmc) is isosceles

Explanation:

Step1: Recall Rectangle Properties

In rectangle \(ABCD\), \(AB = CD\), \(\angle A=\angle D = 90^\circ\), and \(M\) is the midpoint of \(AD\), so \(AM = MD\).

Step2: Prove Triangles Congruent

Consider \(\triangle ABM\) and \(\triangle DCM\). We have \(AB = DC\) (opposite sides of rectangle), \(\angle A=\angle D = 90^\circ\), and \(AM = DM\) (midpoint). By SAS congruence, \(\triangle ABM\cong\triangle DCM\).

Step3: Deduce BM = CM

Since \(\triangle ABM\cong\triangle DCM\), their corresponding sides \(BM\) and \(CM\) are equal.

Step4: Conclude Isosceles Triangle

A triangle with two equal sides (\(BM = CM\)) is isosceles, so \(\triangle BMC\) is isosceles.

Answer:

To prove \(\triangle BMC\) is isosceles given rectangle \(ABCD\) and \(M\) midpoint of \(AD\):

1. Rectangle Properties: \(AB = CD\), \(\angle A=\angle D = 90^\circ\), \(AM = MD\) (midpoint).
2. SAS Congruence: In \(\triangle ABM\) and \(\triangle DCM\), \(AB = DC\), \(\angle A=\angle D\), \(AM = DM\). Thus, \(\triangle ABM\cong\triangle DCM\).
3. Equal Sides: Corresponding sides \(BM = CM\).
4. Isosceles Definition: \(\triangle BMC\) has \(BM = CM\), so it is isosceles.