Line BD is the perpendicular bisector of CA. What additional information would be sufficient to prove that triangle BDA is congruent to triangle BDC by SAS.

DA ≅ DC (since BD is equidistant to A and C)

Explanation:

A line bisector is one that divides a given line into two equal parts. And which is equidistant from the ends of the bisected line.

Given: BD is the perpendicular bisector of CA.

Thus:

BD is equidistant to A and C

Also, BD is a common side to ΔBDA and ΔBDC

DA ≅ DC (since BD is equidistant to A and C)

<BDA ≅ <BDC (congruent property)

<ADB ≅ <DBC (alternate angle property)

<CDB ≅ <ABD (alternate angle property)

Thus,

ΔBDA ≅ ΔBDC (Side-Angle-Side, SAS, congruent property)

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg