Isosceles Triangles

216. Angle Chase: Parallel Lines & a Hidden Isosceles Triangle · Find a, b and c using alternate angles, the exterior angle, and equal base angles

AB ∥ CD with transversal AD, and triangle ACE is isosceles (CA = CE). Drag A: a always equals ∠ADC (alternate angles), b = ∠ADC + ∠DCE (exterior angle), and c = 180° − 2b (isosceles). At the start a = 36°, b = 68°, c = 44°.

This is an angle-chasing problem that ties three results together. Since AB ∥ CD and AD is a transversal, a = ∠DAB equals its alternate angle ∠ADC, so a = 36° (alternate angles). At E, b = ∠AEC is the exterior angle of triangle CDE, so by the exterior-angle theorem it equals the sum of the two remote interior angles: b = 32° + 36° = 68°. The hidden isosceles triangle is ACE (with CA = CE), so its base angles are equal: ∠CAE = ∠AEC = 68°. Finally, by the angle sum of a triangle, c = ∠ACE = 180° − 68° − 68° = 44°.

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Selina ICSE: Isosceles Triangles

What this lesson covers

Try to break it

Try to break it: drag A anywhere along the base. a never stops matching ∠ADC, b stays the exterior-angle sum ∠ADC + ∠DCE, and CA = CE keeps triangle ACE isosceles, so c stays 180° − 2b. The three relationships lock together — the angle-chase cannot fail.

How you build it

Recreate the parallel-line and triangle figure.

Drop point A near the bottom-left.
Drop point B to the right of A, level with it.
Draw the line through A and B.
Drop point C above line AB, toward the left.
Click C, then click line AB, to draw line CD parallel to AB.
Drop point D on the parallel line, far enough to the right that the transversal AD is long. (D must lie outside an arc of radius CA centred at C.)
Draw the segment from A to D.
Compass: click centre C, then click A to set the radius. The arc passes through A and cuts AD again at E.
Click where the arc crosses AD (the crossing that is not A) to mark E — now CE = CA.
Draw the segment from C to E.
Draw the segment from C to A.

The proof, step by step

Prove that the unknown angles are a = 36°, b = 68°, and c = 44°.

AB is parallel to CD, and AD is a transversal. Therefore, alternate interior angles are equal: a = 36°.
In ΔCDE, the exterior angle at E (∠CEA) equals the sum of the two opposite interior angles: 36° + 32° = 68°.
We are given CE = CA, so ΔACE is isosceles. Base angles are equal: b = ∠CEA = 68°.
The angles in ΔACE sum to 180°. So, c = 180° - 68° - 68° = 44°.

Worked example

In the figure, AB || CD and CE = CA. If ∠ADC = 36° and ∠DCE = 32°, what is the value of c?

Using alternate angles, a = 36°. Exterior angle of ΔCDE gives ∠CEA = 36° + 32° = 68°. Since CE = CA, ΔACE is isosceles, so b = 68°. Angle sum in ΔACE gives c = 180° - 68° - 68° = 44°.

36°
44° — correct
68°
72°