<em>Look</em><em> </em><em>at</em><em> </em><em>the</em><em> </em><em>attached</em><em> </em><em>picture</em><em> </em><em>⤴</em>
<em>Hope</em><em> </em><em>it</em><em> </em><em>will</em><em> </em><em>help</em><em> </em><em>u</em><em>.</em><em>.</em>
5x^3/10x^2
*Simplify 5/10 to 1/2
*simplify the exponent so that x^1 is on top
1x/2
FINAL ANSWER:
x/2
Answer:
The proof is below
Step-by-step explanation:
Given a parallelogram ABCD. Diagonals AC and BD intersect at E. We have to prove that AE is congruent to CE and BE is congruent to DE i.e diagonals of a parallelogram bisect each other.
In ΔACD and ΔBEC
AD=BC (∵Opposite sides of a parallelogram are equal)
∠DAC=∠BCE (∵Alternate angles)
∠ADC=∠CBE (∵Alternate angles)
By ASA rule, ΔACD≅ΔBEC
By CPCT(Corresponding Parts of Congruent triangles)
AE=EC and DE=EB
Hence, AE is conruent to CE and BE is congruent to DE
