5. Quadrilaterals

1. In \(\Box\)IJKL, side IJ | | side KL, \(\angle\)I = 108°, \(\angle\)K = 53° then find the measures of \(\angle\)J and \(\angle\)L.

Solution:

side IJ | | side KL ... (Given)
Consider transversal JK
∴ \(\angle\)J + \(\angle\)K = 180° ... (Interior angles)
∴ \(\angle\)J + 53° = 180°
∴ \(\angle\)J = 180° − 53°
∴ \(\angle\)J = 127° ... (i)

Also, in \(\Box\)IJKL,
\(\angle\)J + \(\angle\)K + \(\angle\)L + \(\angle\)I = 360° ... (Theorem)
∴ 127° + 53° + \(\angle\)L + 108° = 360°
∴ \(\angle\)L + 288° = 360°
∴ \(\angle\)L = 360° − 288°
∴ \(\angle\)L = 72° ... (ii)

2. In \(\Box\)ABCD, side BC | | side AD, side AB \(\cong\) side DC If \(\angle\)A = 72° then find the measures of \(\angle\)B, and \(\angle\)D.

Solution:

Draw BE \(\perp\) AD and CF \(\perp\) AD such that A – E – D and A – F – D.

Now, BC | | AD ... (Given)
∴ BE = CF ... (Perpendicular distances between parallel lines) ... (i)

Also, in \(\triangle\)BEA and \(\triangle\)CFD,
\(\angle\)BEA \(\cong\) \(\angle\)CFD ... (90° each – Construction)
seg AB \(\cong\) seg DC ... (Given)
seg BE \(\cong\) seg CF ... [From (i)]
∴ \(\triangle\)BEA \(\cong\) \(\triangle\)CFD ... (Hypotenuse Side Test) ...(ii)
∴ \(\angle\)BAE \(\cong\) \(\angle\)CDF ... (c a c t)
But, \(\angle\)BAE = 72° ... (Given)
∴ \(\angle\)CDF = 72°
i.e. \(\angle\)D = 72° ... (iii)

Also, side BC | | side AD ... (Given)
Consider transversal AB
∴ \(\angle\)A + \(\angle\)B = 180° ... (Interior angles)
∴ 72° + \(\angle\)B = 180° ... (Given)
∴ \(\angle\)B = 180° − 72°
∴ \(\angle\)B = 108° ... (iv)

3. In \(\Box\)ABCD, side BC < side AD (Figure 5.32), side BC | | side AD and if side BA \(\cong\) side CD then prove that \(\angle\)ABC \(\cong\) \(\angle\)DCB.

Proof:

Draw BE \(\perp\) AD and CF \(\perp\) AD such that A – E – D and A – F – D.

Now, BC | | AD ... (Given)
∴ BE = CF ... (Perpendicular distances between parallel lines) ... (i)

Also, in \(\triangle\)BEA and \(\triangle\)CFD,
\(\angle\)BEA \(\cong\) \(\angle\)CFD ... (90° each – Construction)
seg AB \(\cong\) seg DC ... (Given)
seg BE \(\cong\) seg CF ... [From (i)]
∴ \(\triangle\)BEA \(\cong\) \(\triangle\)CFD ... (Hypotenuse Side Test) ...(ii)
∴ \(\angle\)BAE \(\cong\) \(\angle\)CDF ... (c a c t)
i.e. \(\angle\)BAD \(\cong\) \(\angle\)CDA ... (ii)
... (∵ A – E – D and A – F – D)

Now, side BC | | side AD ... (Given)
Consider transversal AB
\(\angle\)ABC + \(\angle\)BAD = 180° ... (Interior angles) ... (iii)
Consider transversal CD
\(\angle\)DCB + \(\angle\)CDA = 180° ... (Interior angles) ... (iv)

From (iii) and (iv), we have
\(\angle\)ABC + \(\angle\)BAD = \(\angle\)DCB + \(\angle\)CDA
But, \(\angle\)BAD \(\cong\) \(\angle\)CDA ... [From (ii)]
∴ \(\angle\)ABC = \(\angle\)DCB ... (v)