Beautifully Crafted Circles

Two circles Γ and Σ in the plane intersect at two distinct points A and B and the centre of Σ lies on Γ . Let points C and D be on Γ and Σ, respectively, such that C, B and D are collinear. Let point E on Σ be such that DE is parallel to AC. Show that AE = AB.

#Geometry

Easy Math Editor

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Comments

Let $O$ be the centre of Γ, and let $P$ be the centre of Σ. Let the point $F$ be on Γ, such that $F,O,P$ are collinear. Let $\angle ACB = x$ .

From the picture, we can see that there are two similar possibilities for how the diagram turns out. By properties of parallel lines, $\angle BDE$ either equals $x$ or $180^{\circ} - x$ . In both cases, since $ABDE$ is a cyclic quadrilateral, this implies that $\angle BAE = x = \angle ACB$ . [1]

Since $FCBA$ is a cyclic quadrilateral, we have $\angle ACB = \angle AFB$ . Combining this with [1] gives us $\angle BAE = \angle BFA$ . [2]

Since $F,O,P$ are collinear, then $FP$ is a diameter of Γ. Therefore $\angle FAP = \angle FBP = 90^{\circ}$ . [3]

Since $PA$ and $PB$ are radii of Σ, we have $PA = PB$ . Therefore the triangles $\triangle FPA$ and $\triangle FPB$ have two pairs of sides equal ( $PA = PB$ and $FP = FP$ ) and one pair of right angles. Hence $\triangle FPA$ and $\triangle FPB$ are congruent. Hence $FA = FB$ . [4]

Since $P$ is the center of Σ and $E$ is a point on Σ on the same side of $AB$ as $P$ , therefore $\angle BPA = 2\angle BEA$ . Now since $\triangle BPA$ is an isosceles triangle, and since the angles in a triangle sum to $180^{\circ}$ , we have $\angle BAP = \dfrac{180^{\circ} - \angle BPA}{2} = 90^{\circ} - \angle BEA$ . But thanks to [3], we know that $\angle FAB = 90^{\circ} - \angle BAP$ . Hence, $\angle FAB = \angle BEA$ . [5]

Therefore, by [2] and [5], the triangles $\triangle BEA$ and $\triangle FAB$ have two pairs of equal angles, which means they have three pairs of equal angles. Therefore $\triangle BEA$ and $\triangle FAB$ are similar triangles.

Therefore, by [4], since $FA = FB$ , then $AE = AB$ also. QED

Ariel Gershon - 3 years, 11 months ago

thats a very good solution you are the best

Shashwat Gokhe - 3 years, 11 months ago

Thanks! This was a very interesting problem.

Ariel Gershon - 3 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$