Pure Angle Chasing
What Might This Be About?
Problem
Circles $(O)$ and $(Q)$ intersect in $A$ and $B;$ $C$ on $(O);$ $AC,$ $BC$ meet $(Q)$ second time in $D$ and $E,$ respectively.
Prove that $OC$ is perpendicular to $DE.$
Solution 1
This is a quintessential "angle chasing".
We use properties of inscribed angles. Let $OC$ meet $(Q)$ in $F$ and $G$ and $(O)$ in $K$ as shown. Let $H$ be the intersection of $OC$ and $DE.$ Then
$\displaystyle \angle CDH=\angle ADE=\angle ABE=\angle ABC=\frac{1}{2}\overparen{AC},$
$\displaystyle \angle DCH=\angle ACK=\frac{1}{2}\overparen{AK}.$
It follows that $\displaystyle\angle CDH+\angle DCH=\frac{1}{2}\overparen{CK}=90^{\circ}.$ But then (from $\Delta CDH)$ $\angle CHD =180^{\circ}-(\angle CDH+\angle DCH)=90^{\circ}.$
Solution 2
Machò Bònis posted another solution at the Cut the Knot facebook page:
Draw a tangent to $(O)$ at $C.$ The result follows immediately by a limiting case of Reim's theorem.
Acknowledgment
I borrowed this problem from Antonio Gutierrez after he broadcasted it at tweeter.com.
Chasing Inscribed Angles
- Munching on Inscribed Angles
- More On Inscribed Angles
- Inscribed Angles
- Tangent and Secant
- Angles Inscribed in an Absent Circle
- A Line in Triangle Through the Circumcenter
- Angle Bisector in Parallelogram
- Phantom Circle and Recaptured Symmetry
- Cherchez le quadrilatere cyclique
- Cyclic Quadrilateral, Concurrent Circles and Collinear Points
- Parallel Lines in a Cyclic Quadrilateral
- Reim's Similar Coins I
- Reim's Similar Coins II
- Reim's Similar Coins III
- Reim's Similar Coins IV
- Pure Angle Chasing
- Pure Angle Chasing II
- Pure Angle Chasing III
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