To prove: OA + OB + OC+OD > AC + BD
Construction:Join OA, OB, OC and OD. Also, join AC and BD
Proof : Consider ΔBOD, by triangle in equality (sum of any two sides of a triangle is greater than the third side),
We have OB + OD >BD .....(1)
Similarly, in ΔAOC, by triangle inequality, OA + OC > AC .... (2)
Adding (1) and (2)
(OB + OD) + OA + OC ) > BD + AC
⇒ OA + OB + OC + OD > AC + BD
Hence, the result is proved.
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Here is the detailed explanation.
Let O be any point within the quadrilateral ABCD.
To prove: OA + OB + OC > AC + BD
Construction:Join OA, OB, OC and OD. Also, join AC and BD
Proof : Consider ΔBOD, by triangle in equality (sum of any two sides of a triangle is greater than the third side),
We have OB + OD >BD .....(1)
Similarly, in ΔAOC, by triangle inequality, OA + OC > AC .... (2)
Adding (1) and (2)
(OB + OD) + OA + OC ) > BD + AC
⇒ OA + OB + OC + OD > AC + BD
Hence, the result is proved.
a and d
never
\8
What is ABCD? It is a quadrilateral, but what it is? What is 3x - 7? A side, a perimeter, or what else? Please, give me some additional information in order to be able to solve this problem.
A rectangle.
You cannot prove it since it is not true for a general quadrilateral.
none of these answers are correct
This is the image
A quadrilateral is a parallelogram if one pair of opposite sides are equal and parallel Let ABCD be a quadrilateral in which ABCD and AB=CD, where means parallel to. Construct line AC and create triangles ABC and ADC. Now, in triangles ABC and ADC, AB=CD (given) AC = AC (common side) Angle BAC=Angle ACD (corresponding parts of corresponding triangles or CPCTC) Triangle ABC is congruent to triangle CDA by Side Angle Side Angle BCA =Angle DAC by CPCTC And since these are alternate angles, ADBC. Thus in the quadrilateral ABCD, ABCD and ADBC. We conclude ABCD is a parallelogram. var content_characters_counter = '1032';
-+-
a and d
never
never
"abcd is not a parallelogram or it does not have any right angles." ~(P and Q) = ~P or ~Q
always
\8
none of these are correct