Isosceles Triangles

217. Angle Chase: Nested Isosceles Triangles · two given isosceles triangles force ∠BAD : ∠ADB into a 3 : 1 ratio

CA = CD and AB = BC make triangles ACD and ABC isosceles, so ∠BAD is always exactly three times ∠ADB.

CDB∠BAD = 60°∠BAD = 60°∠ADB = 20°∠ADB = 20°CA = 200CA = 200CD = 200CD = 200AB = 131AB = 131BC = 131BC = 131∠BAD ÷ ∠ADB = 3∠BAD ÷ ∠ADB = 3A
This is a worked angle chase — an example, not a new theorem. The figure is built from two given isosceles triangles: in △ACD we are told CA = CD, and in △ABC we are told AB = BC. (The circle, centred at C through A and D, is just a way to keep CA = CD as you drag — and B is pinned so that AB = BC stays true.) Now follow the angles. Let ∠ADB = x. Equal base angles of △ACD give ∠CAD = x; the exterior angle of △ACD at C gives ∠ACB = 2x; equal base angles of △ABC give ∠BAC = 2x; so ∠BAD = ∠BAC + ∠CAD = 3x. Hence ∠BAD : ∠ADB = 3 : 1 at every position — drag A to check.

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Selina ICSE: Isosceles Triangles

What this lesson covers

Try to break it

Drag A around the circle. The two isosceles triangles ABC and ACD share side AC, and the relationship ∠BAD = 3 × ∠ADB holds at every position. Try to drag A so the ratio drifts away from 3:1 — impossible. The double-isosceles construction locks it in.

How you build it

Construct the two nested isosceles triangles: △ABC with AB = BC and △ACD with CA = CD.

  • Point tool: mark point B near the top — the left end of the base.
  • Point tool: mark point C to the right of B and level with it.
  • Line tool: click B then C. C, and later D, sit on this base line — it extends right of C for D.
  • Arc tool: click B, then C. The arc has radius BC — every point on it is exactly BC from B.
  • Point tool: mark the apex A on the lower part of the arc, below the base line — so the triangle hangs downward to match the figure above. Because A is on the arc, BA = BC, so triangle ABC is isosceles.
  • Segment tool: join B to A.
  • Segment tool: join C to A.
  • Arc tool: click C, then A. This arc has radius CA — it carries the length CA down onto the base.
  • Point tool: mark D where this arc crosses the base line, to the right of C. Now CD = CA, so triangle ACD is isosceles.
  • Segment tool: join A to D. With triangle ABC isosceles (BA = BC) and triangle ACD isosceles (CA = CD), the angle chase gives ∠BAD = 3 × ∠ADB.

The proof, step by step

Prove that ∠BAD : ∠ADB is exactly 3 : 1.

  • Let ∠ADB = x. In ΔACD, AC = CD, so the base angles are equal: ∠CAD = ∠CDA = x.
  • The exterior angle ∠ACB of ΔACD equals the sum of the two opposite interior angles: ∠ACB = ∠CAD + ∠CDA = x + x = 2x.
  • In ΔABC, AB = BC, so the base angles are equal: ∠BAC = ∠ACB = 2x.
  • Now, ∠BAD = ∠BAC + ∠CAD = 2x + x = 3x.
  • Therefore, ∠BAD : ∠ADB = 3x : x = 3 : 1. Hence Proved.

Worked example

In the given figure, AB = BC and AC = CD. If ∠ADB = 18°, find the measure of ∠BAD.

Since AC = CD, ∠CAD = ∠CDA = 18°. Exterior ∠ACB = 18° + 18° = 36°. Since AB = BC, ∠BAC = ∠ACB = 36°. Thus, ∠BAD = ∠BAC + ∠CAD = 36° + 18° = 54°.

  • 36°
  • 45°
  • 54° — correct
  • 72°
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