Prove that diagonals divides a parallelogram into 2 triangles of equal area
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Answer:
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Step-by-step explanation:
Here, ABCD is a parallelogram and BD is diagonal.
In △ADB and △CBD
AD∥BC
⇒ ∠ADB=∠CBD [ Alternate angles ]
Also AB∥DC
⇒ ∠ABD=∠CDB [ Alternate angles ]
⇒ DB=BD [ Common side ]
∴ △ADB≅△CBD [ By SAS congruence rule ]
Since congruent figures have same area.
∴ ar(ADB)=ar(CBD) [ Hence, proved ]
∴ We have proved that a diagonal divides a parallelogram into two triangles of equal area.
Answer:
A diagonal divides a parallelogram into two triangles of equal area. ...
A diagonal divides a parallelogram into two triangles of equal area. ...angle abd. = angle bdc (Alternate interior.angles)
A diagonal divides a parallelogram into two triangles of equal area. ...angle abd. = angle bdc (Alternate interior.angles)angle dbc =angle cdb. (")
A diagonal divides a parallelogram into two triangles of equal area. ...angle abd. = angle bdc (Alternate interior.angles)angle dbc =angle cdb. (")
A diagonal divides a parallelogram into two triangles of equal area. ...angle abd. = angle bdc (Alternate interior.angles)angle dbc =angle cdb. (") and a common side of bd
A diagonal divides a parallelogram into two triangles of equal area. ...angle abd. = angle bdc (Alternate interior.angles)angle dbc =angle cdb. (") and a common side of bd Since congruent figures have same area. ∴ ar(ADB)=ar(CBD) [ Hence, proved ] ∴ We have proved that a diagonal divides a parallelogram into two triangles of equal area.
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