Given: ΔABC isosceles
and the line AD cuts the side BC.
Are ΔADC and ΔADB congruent?
And if so, according to which congruence theorem?
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Given: ΔABC isosceles
and the line AD cuts the side BC.
Are ΔADC and ΔADB congruent?
And if so, according to which congruence theorem?
Since we know that the triangle is isosceles, we can establish that AC=AB and that
AD=AD since it is a common side to the triangles ADC and ADB
Furthermore given that the line AD intersects side BC, we can also establish that BD=DC
Therefore, the triangles are congruent according to the SSS (side, side, side) theorem
Congruent by L.L.L.
Look at the triangles in the diagram.
Which of the statements is true?
You need additional information to conclude BD = DC! The diagram alone doesn't guarantee this. If AD is the perpendicular bisector or if D is the midpoint of BC, then BD = DC.
For SAS, you need two sides and the included angle. While we have AD = AD (common side) and potentially AC = AB (isosceles), we need the angles ∠CAD and ∠BAD to be equal, which isn't automatically true.
The isosceles property gives us AC = AB automatically, but we still need to establish the third pair of equal sides (BD = DC) through additional given information or proof.
Yes! If AD is just any line cutting BC (not perpendicular or creating a midpoint), then BD ≠ DC and the triangles won't be congruent by SSS. The position of point D matters!
Count what you know and match the pattern!
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