D and E are the mid-points of the sides AB and AC of ΔABC and O is any point on side BC. O is joined to A. If P and Q are the mid-points of OB and OC
respectively, then prove that DEQP is a parallelogram.

Check out the diagram below.

Since D and E are midpoints of AB and AC this means midsegment DE is parallel to side BC. Furthermore, it means DE is parallel to PQ because PQ is entirely on BC.

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Note how I drew a line from A to O. This forms two triangles ABO and ACO

For triangle ABO, segment DP is a midsegment because D and P are midpoints of AB and OB respectively. This means side DP is parallel to side AO. Keep this in mind as we'll use it later.

For triangle ACO, segment EQ is parallel to AO because EQ is a midsegment for similar reasons as already mentioned (E and Q are midpoints of AC and OC respectively). So this must mean that AO is parallel to EQ

Now chain the arguments together: if DP || AO and AO || EQ, then DP || EQ by the transitive property.

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In short, we've proven two things

This is enough to conclude that DEQP is a parallelogram. By definition, a parallelogram has two pairs of opposite parallel sides.

the equation cos(35) =a/25 can be used to find the length of bc what is the length of bc? round to the nearest tenth

step-by-step explanation:

16.80 times .20= 3.36

step-by-step explanation: