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)

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

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