Question

In the adjoining figure , APB and AQC are equilateral triangles. Prove that PC = BQ. ( Hint : %20%5Ctriangle" id="TexFormula1" title=" \triangle" alt=" \triangle" align="absmiddle" class="latex-formula">APC = $\triangle$AQB, then PC = BQ)
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Answer 1

Answer:

See Below.

Step-by-step explanation:

Statements:                                                           Reasons:

$\displaystyle 1)\text{ } \Delta APB \text{ and } \Delta AQC \text{ are equilateral triangles}$      Given

$\displaystyle 2) \text{ } m \angle PAB = 60$                                                     Definition of equilateral.

$3)\text{ } m \angle QAC = 60$                                                     Definition of equilateral.

$4)\text{ } m\angle PAB = m\angle QAC$                                          Substitution

$5)\text{ } m\angle PAC=m\angle PAB+m\angle BAC$                       Angle Addition

$\displaystyle 6)\text{ } m\angle QAB=m\angle QAC+m\angle BAC$                       Angle Addition

$7)\text{ } m\angle QAB=m\angle PAB+m\angle BAC$                       Substitution

$\displaystyle 8)\text{ } m\angle PAC=m\angle QAB$                                         Substitution

$9)\text{ } PA=BA$                                                          Definition of equilateral

$10)\text{ } AC=AQ$                                                        Definition of equilateral

$\displaystyle 11)\text{ } \Delta PAC \cong \Delta BAQ$                                            Side-Angle-Side Congruence*

$\displaystyle 12)\text{ } PC=BQ$                                                        CPCTC

* SAS Congruence:

PA = BA

∠PAC = ∠QAB

AC = AQ

Answer 2

Answer:

In ΔAPB and ΔAQC,

PA = PB

anglePAC = angleQAB

AC = AQ

therefore, ΔAPC ≅ ΔAQB by SAS congruency.

So by CPCT,

PC = PQ. proved..