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given ac≅bc & ad≅bd.
prove ∆adc≅∆bdc

I have dyscalculia and no matter how many times they go thru this I just cannot figure out proofs.

Given Acbc Amp Adbd Prove Adcbdc I Have Dyscalculia And No Matter How Many Times They Go Thru This I Just Cannot Figure Out Proofs class=

Sagot :

Answer:

[tex]\begin{array}{|c|c|}\cline{1-2} \sf Statements & \sf Reasons \\\cline{1-2} \phantom{\dfrac12}\overline{AC} \cong \overline{BC} & \sf Given \\\cline{1-2} \phantom{\dfrac12}\overline{AD} \cong \overline{BD} & \sf Given \\\cline{1-2} \phantom{\dfrac12}\overline{AD} \cong \overline{AD} & \sf Common \\\cline{1-2} \phantom{\dfrac12}\triangle ADC \cong \triangle BDC & \sf SSS \\\cline{1-2} \end{array}[/tex]

Step-by-step explanation:

Congruence sign

≅ sign means congruent, i.e. they both have the same shape and the same size.

SSS (Side-Side-Side) Congruence Theorem

If all three sides of one triangle are equal to all three corresponding sides of another triangle, then the two triangles are congruent.

Proof

We are told that AC and BC are equal, and that AD and BD are equal. Therefore, two of the three sides of the two triangles are congruent.

As the two triangles share side AD in common, this means that all three corresponding sides of both triangles are equal, and therefore the two triangles are congruent according to the SSS theorem.

[tex]\begin{array}{|c|c|}\cline{1-2} \sf Statements & \sf Reasons \\\cline{1-2} \phantom{\dfrac12}\overline{AC} \cong \overline{BC} & \sf Given \\\cline{1-2} \phantom{\dfrac12}\overline{AD} \cong \overline{BD} & \sf Given \\\cline{1-2} \phantom{\dfrac12}\overline{AD} \cong \overline{AD} & \sf Common \\\cline{1-2} \phantom{\dfrac12}\triangle ADC \cong \triangle BDC & \sf SSS \\\cline{1-2} \end{array}[/tex]