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ABC is a triangle - Leaving Cert Mathematics - Question 6B - 2014

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Question 6B

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ABC is a triangle. D is the point on BC such that AD ⊥ BC. E is the point on AC such that BE ⊥ AC. AD and BE intersect at O. Prove that $|∠DOC| = |∠DEC|$.

Worked Solution & Example Answer:ABC is a triangle - Leaving Cert Mathematics - Question 6B - 2014

Step 1

Consider the quadrilateral DOEC.

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Answer

We know that the angles in quadrilateral DOEC satisfy the following:

CDO+EOC+OEC+ODC=180°.|∠CDO| + |∠EOC| + |∠OEC| + |∠ODC| = 180°.

Given that:

  • CDO=90°|∠CDO| = 90° (since AD ⊥ BC)
  • EOC=90°|∠EOC| = 90° (since BE ⊥ AC)

Thus, substituting these values gives us:

CDO+EOC=90°+90°=180°.|∠CDO| + |∠EOC| = 90° + 90° = 180°.

This shows both pairs of opposite angles sum to 180°. Therefore, DOEC is a cyclic quadrilateral.

Step 2

Prove that |∠DOC| = |∠DEC|.

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Answer

Since DOEC is a cyclic quadrilateral, the opposite angles are equal according to the properties of cyclic quadrilaterals.

Therefore, we conclude that:

DOC=DEC.|∠DOC| = |∠DEC|.

This completes the proof as required.

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