A parallelogram is a quadrilateral with opposite sides parallel (and therefore opposite angles equal). A quadrilateral with equal sides is called a rhombus, and a parallelogram whose angles are all right angles is called a rectangle.
\begin{equation}
\begin{array}{l}
{\text { 4. Let } A B C D \text { be a parallelogram. Let } O \text { be a point in its interior such that } \angle A O B+} \\
{\angle D O C=180^{\circ} . \text { Show that } \angle O D C=\angle O B C \text { . }}
\end{array}
\end{equation}
CMI Entrance, 2019
Parallelogram
6 out of 10
Challenging Problems in Geometry
First construct AP and BP parallel to DO and CO respectively. Basically translate triangle DOC to triangle APB with DC and AB as parallel bases respectively. Notice that product of the fractions is 1. Can you use this fact to compute the geometric mean of the fractions?
Now using angle manipulations by spotting the required parallelograms and show that APBO is cyclic quadrilateral.
Now using Hint 2 and the fact that PBCO is a parallelogram arrive at the proof.
After implementing Hint 3 by spotting the parallelogram PBCO, notice that the the diagonal OB forms the the alternate interior angles i.e. angle POB and angle OBC which will be equal to each other as PB is parallel to OC since PBCO is a parallelogram. Since angle POB is equal to angle PAB (angle POB and angle PAB share the same arc AB and hence are equal to each other), which again itself is equal to angle ODC as AP is parallel to DO as we translated triangle DOC to triangle APB. Hence, angle ODC = angle PAB = angle POB = angle OBC. Hence we are done.
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