ABCD is a rhombus and RD = CS. SCB, DCR and AQR are straight lines. Prove that triangles SCD and RDA are congruent.

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Worked solution

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1. Identify equal corresponding sides from the given condition and the rhombus property.

Given that

RD=CSRD = CS.

Since ABCDABCD is a rhombus, all its sides are equal. Hence

CD=DACD = DA.

2. Show the included angles are equal using collinearity and parallel sides of the rhombus.

Since S,C,BS,C,B are collinear and AD∥CBAD \parallel CB, we have

CS∥ADCS \parallel AD.

Since D,C,RD,C,R are collinear, CDCD and RDRD lie on the same straight line.

Therefore,

∠SCD=∠RDA\angle SCD = \angle RDA.

3. Conclude triangle congruence by SAS.

In △SCD\triangle SCD and △RDA\triangle RDA,

CS=RDCS = RD, CD=DACD = DA, and ∠SCD=∠RDA\angle SCD = \angle RDA.

Hence

△SCD≅△RDA(SAS)\triangle SCD \cong \triangle RDA \quad \text{(SAS)}.

Answer

Since ABCDABCD is a rhombus, CD=DACD = DA and AD∥CBAD \parallel CB. Given that RD=CSRD = CS. Also, S,C,BS,C,B are collinear and D,C,RD,C,R are collinear, so CS∥ADCS \parallel AD and CDCD lies on the same straight line as DRDR. Hence

∠SCD=∠RDA\angle SCD = \angle RDA.

Therefore, in △SCD\triangle SCD and △RDA\triangle RDA,

CS=RDCS = RD, CD=DACD = DA, and ∠SCD=∠RDA\angle SCD = \angle RDA.

Thus △SCD≅△RDA\triangle SCD \cong \triangle RDA by SAS.

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