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
4(c-d) is the answer bc the difference of (c-d) is done first, then multiplied by 4
Answer:hard to do without a pic
is there a pic of the triangle
Step-by-step explanation:
For example, exc. 15.
x + y = 10 and 6/x = 9/y; You need y!
=> x = 10 -y and 9x = 6y => 9(10-y) = 6y => 90 - 9y = 6y => 15y = 90 => y = 6.
Answer:
im pretty sure its 1.4
Step-by-step explanation:
