Question

Abcd is a trapezium in which ab parallel to dc then find ac^2 + bd^2

Answer 1

Answer:

Step-by-step explanation:

In a trapezium ABCD, if AB||CD, then prove that AC²+BC²=BC²+AD²+2AB*C.

In a trapezium ABCD, if AB||CD, then AC2+BC2 is equal to?

Edit: the new version of the question is false: the left hand side should be AC²+BD².

In a trapezium ABCD, if AB||CD, then prove that AC²+BC²=BC²+AD²+2AB*CD?

I previously showed in my answer to the original question below that for an isosceles trapezium, AC2+BD2=BC2+AD2+2AB×CD .

I will edit it below to investigate the general trapezium.

In a trapezium ABCD, if AB||CD, then AC2+BC2 is equal to?

I presume you mean AC2+BC2. Are the vertices labelled consecutively around the figure or are the parallel sides labelled from left to right (or right to left)? In the former case it would be better to write AB||DC and AC and BD would be the diagonals. In the latter case AC and BD would be two sides.

Use the cosine rule in traingles ABC, ACD, ABD and BCD:

AC2=AB2+BC2−2AB\timesBCcosB , AC2=AD2+CD2−2AD×CDcosD, BD2=AB2+AD2−2AB×ADcosA, BD2=BC2+CD2−2BC×CDcosC

therefore

AB(AC2−AD2−CD2)=−CD(BD2−AB2−AD2), CD(AC2−AB2−BC2)=−AB(BD2−BC2−CD2)

so

AC2−AD2CD+BD2−AD2AB=AB+CD, AC2−BC2AB+BD2−BC2CD=AB+CD

thus

ABAC2+CDBD2=(AD2+AB×CD)(AB+CD), CDAC2+ABBD2=(BC2+AB×CD)(AB+CD)

and finally,

AC2+BD2=AD2+BC2+2AB×C

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