Question

In the adjoining figure , APB and AQC are equilateral triangles. Prove that PC = BQ. ( Hint :
`\triangle`APC =
`\triangle`AQB, then PC = BQ)

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Answer 1

See Below.

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