Given: BC ≅ CD and AC bisects ∠BCD Prove: ∆ABC ≅ ∆ADC

weshjfg

Step-by-step explanation:

"standard form:

3x2 + 14x − 10 = 0

solutions based on quadratic formula:

x1   = −14 − √ 142 − 4×3×(−10) /   2×3   = −14 − 2 × √ 79 /   6   ≈ −5.296

x2   = −14 + √ 142 − 4×3×(−10) /   2×3   = −14 + 2 × √ 79 /   6   ≈ 0.629

dont't really have enough info here but its likely

sss for the three sides being congruent ( its stands for side side side)

or

sas   for side angle side.   so a side is congruent and and angle and a side