Proof:
A rhombus is a parallelogram such that AB=DC=AD=BC ---(i) Also the diagonals of a parallelogram bisect each other. Hence BO=DO and AO=OC ---(ii) Now compare triangles AOB and AOD, AB=AD (from (i) above) BO=DO (from (ii) above) AO=AO (common side) (SSS congruency condit..
A rhombus is a parallelogram such that AB=DC=AD=BC ---(i) Also the diagonals of a parallelogram bisect each other. Hence BO=DO and AO=OC ---(ii) Now compare triangles AOB and AOD, AB=AD (from (i) above) BO=DO (from (ii) above) AO=AO (common side) (SSS congruency condit..Proof:
) i.e., the diagonals of a parallelogram bisect each other. We have proved one of the properties of a parallelogram by logical deductio..
Proof:
In the diagram if D is the mid-point of AB and DE is drawn parallel to BC, then E Will be the midpoint of AC i.e., AE=EC. Now if AX is drawn parallel to BC, then also AE=EC if AD=DB. Now draw three parallel lines AB, CD, EF as shown in the figure. Draw a transversal t 1 such that AB=BC. N..
In the diagram if D is the mid-point of AB and DE is drawn parallel to BC, then E Will be the midpoint of AC i.e., AE=EC. Now if AX is drawn parallel to BC, then also AE=EC if AD=DB. Now draw three parallel lines AB, CD, EF as shown in the figure. Draw a transversal t 1 such that AB=BC. N..Proof:
Now in triangles, ABD and ABC, AB=AB (common side) (each angle is a right angle) AD=BC (opposite sides of parallelogram) BD=AC (corresponding parts of corresponding triangles) Hence the theorem is prov..
Now in triangles, ABD and ABC, AB=AB (common side) (each angle is a right angle) AD=BC (opposite sides of parallelogram) BD=AC (corresponding parts of corresponding triangles) Hence the theorem is prov..Proof:
In triangles ABC and ADC, AB=CD (given) AC = AC (common side) (corresponding parts of corresponding triangles) Since these are alternate angles, AD||BC. Thus in the quadrilateral ABCD, AB||CD and AD||BC. ABCD is a parallelogram..
In triangles ABC and ADC, AB=CD (given) AC = AC (common side) (corresponding parts of corresponding triangles) Since these are alternate angles, AD||BC. Thus in the quadrilateral ABCD, AB||CD and AD||BC. ABCD is a parallelogram..Proof:
In triangles ABC and DBC, AB=DC (opposite sides of parallelogram) BC=BC (common side) AC=BD (given) (corresponding parts of corresponding triangles) But these angles are consecutive interior angles on the same side of transversal BC and AB||DC. By definition of rectangle, parallelogr..
In triangles ABC and DBC, AB=DC (opposite sides of parallelogram) BC=BC (common side) AC=BD (given) (corresponding parts of corresponding triangles) But these angles are consecutive interior angles on the same side of transversal BC and AB||DC. By definition of rectangle, parallelogr..Proof:
In triangles AOB and COD, AO = CO (given) BO = OD (given) (vertically opposite angles are equal) D AOB D COD (SAS congruency condition) Since these are alternate angles made by the transversal AC intersecting AB and CD. AB||CD Similarly, AD||BC Hence ABCD is a parallelogram..
In triangles AOB and COD, AO = CO (given) BO = OD (given) (vertically opposite angles are equal) D AOB D COD (SAS congruency condition) Since these are alternate angles made by the transversal AC intersecting AB and CD. AB||CD Similarly, AD||BC Hence ABCD is a parallelogram..Proof:
..
..Proof:
(i) ( Theorem) From (i) and (ii), ..
(i) ( Theorem) From (i) and (ii), ..Proof:
( Linear pair) ( corresponding angles axiom) Similarly, we can prove that..
( Linear pair) ( corresponding angles axiom) Similarly, we can prove that..See what our Users say :
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