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given: \\( \\overline { f g } \\cong \\overline { f h } \\), \\( \\over…

Question

given: \\( \overline { f g } \cong \overline { f h } \\), \\( \overline { f k } \\) bisects \\( \angle h f g \\). prove: \\( \triangle g j k \cong \triangle h j k \\) 6. 7. \\( \angle g j f \\) and \\( \angle g j k \\) are supplementary; \\( \angle h j f \\) and \\( \angle h j k \\) are supplementary. 8. \\( \angle g j k \cong \angle h j k \\) 6. corresponding parts of congruent tri - angles are congruent (c.p.c.t.c.) 7. if two angles are a linear pair, then they are supplementary. 8.

Explanation:

Step1: Use the definition of angle bisector

Since \( \overline{FK} \) bisects \( \angle HFG \), then \( \angle GFK=\angle HFK \).

Step2: Use the Side - Angle - Side (SAS) congruence criterion

We know that \( \overline{FG}\cong\overline{FH} \) (given), \( \angle GFK = \angle HFK \) (from step 1), and \( \overline{FK}\cong\overline{FK} \) (common side). So, \( \triangle FGK\cong\triangle FHK \) by SAS.

Step3: Use the Corresponding Parts of Congruent Triangles are Congruent (CPCTC)

Since \( \triangle FGK\cong\triangle FHK \), then \( \overline{GK}\cong\overline{HK} \) and \( \angle FGK=\angle FHK \).

Step4: Consider \( \angle GJF \) and \( \angle GJK \) are supplementary; \( \angle HJF \) and \( \angle HJK \) are supplementary (given)

Let \( \angle GJF = x\), then \( \angle GJK=180 - x\). Let \( \angle HJF = y\), then \( \angle HJK = 180 - y\).

Step5: Use the fact that \( \angle FGK=\angle FHK \) (from step 3) and angles in triangles

In \( \triangle GJF\) and \( \triangle HJF\), we can find relations for angles. But since \( \triangle FGK\cong\triangle FHK \), and considering the linear - pair supplementary angles.

Step6: Use the Side - Side - Side (SSS) or Side - Angle - Side (SAS) for \( \triangle GJK\) and \( \triangle HJK\)

We have \( \overline{GK}\cong\overline{HK} \) (from step 3), \( \overline{JK}\cong\overline{JK} \) (common side), and \( \angle GJK\cong\angle HJK \) (from step 4 and angle relations). So, \( \triangle GJK\cong\triangle HJK \) by SAS.

Answer:

\( \triangle GJK\cong\triangle HJK \)