Sequential hints for geometry problems are useful. It is much better than ‘reading solutions’. Use this interesting problem from Indian National Math Olympiad (INMO 2019) to refresh your geometry skills.

**Angle Chasing can be one of the most used tools in geometry. But it nevertheless requires some additional insight to finish off the problem.**

## Problem

Let \(ABC\) be a triangle with \(\angle BAC > 90^\circ \). Let \(D\) be a point on the line segment \(BC\) and \(E\) be a point on the line \(AD\) such that \(AB\) is tangent to the circumcircle of triangle \(ACD\) at \(A\) and \(BE\) perpendicular to \(AD\). Given that \(CA=CD\) and \(AE=CE\), determine \(\angle BCA\) in degrees

### Hint 1: Draw a picture!

Also see

Math Olympiad Program at Cheenta

## Hint 2: Alternate Segment Theorem

There is a relation between tangent and chord (and the angles they create). Here you go:

Crux of the theorem: Suppose BA is tangent to a circle at the point A. From A, if we draw a chord, then two angles are created:

- \( \angle BAD \) = angle between the tangent and chord
- \( \angle DCA \) = angle subtended (created) by the chord DA at any point C in the
**alternate arc DA**

These two angles are equal (Try to prove this!)

**What is ‘alternate’? **

Notice that the smaller arc of the circle intersects with \( \angle BAD \). Alternate arc DA implies we have to consider the **other** arc (in this case the bigger arc DA) and choose a point C from it.

Use the alternate segment theorem to conclude: \( \angle BAD = \angle ACD \)

## Hint 3: Draw the circumcircle

This is a major hint!

Ready?

Draw the circumcircle of ABC.

You should definitely stop reading at this point and try the problem on your own.

## Hint 4: What is the position of E?

Extend AE to meet the circumference of ABC at F.

Can you show that E is the center of this circle?

## Hint 5: E is center

\(\angle AED = \angle ACD \) because both of them are subtended by the same arc AB at the circumference.

Hence ABF is isosceles because (\(\\angle AED = \angle ACD =\angle BAF \)

As BE is perpendicular to the base of the isosceles triangle, it bisects the base implying AE = EF.

We already know AE = CE.

Hence A, C, F are equidistant from E. Hence E is the center.

## Hint 6: Final angle chasing.

Since AF is diameter, \( \angle ABF = 90^0 \) (angle in the semicircle).

Thus \( \angle BAF = 45^0 \) implying \( \angle BCA = 45^o \)

## Reference Book

Challenges and Thrills of Pre-college Mathematics.