Congruency: Congruent Triangles
132. The Isosceles Mirror · RHS congruence in action
ΔAOB and ΔAOC are always congruent right triangles.
Worked proof: when AB = AC and AO ⊥ BC, triangles ΔAOB and ΔAOC are congruent by RHS (Right-angle Hypotenuse Side). The hypotenuse AB = AC is given; AO is common. So OB = OC by CPCT.
What this lesson covers
Try to break it
Drag B left or right along the base. C mirrors it so that AB = AC. Try to make △AOB and △AOC look different — impossible. The perpendicular from A to BC always splits an isosceles triangle into two congruent halves.
How you build it
Construct an isosceles triangle with altitude.
- Mark point B — the left end of the base.
- Mark point C — the right end of the base.
- Draw the base segment BC.
- Mark the midpoint O of BC — click B, then C; the midpoint tool drops O exactly halfway.
- At O, construct the perpendicular to BC going upward — click O first, then a point above BC.
- Mark point A anywhere on the perpendicular — this is the apex of the isosceles triangle.
- Draw segment AB — one equal side of the triangle.
- Draw segment AC — the other equal side. Because A lies on the perpendicular bisector of BC, AB = AC automatically, and ∠AOB = ∠AOC = 90°.
The proof, step by step
Prove that triangles AOB and AOC are congruent.
- In ΔAOB and ΔAOC, AB = AC (Given)
- AO = AO (Common side)
- ∠AOB = ∠AOC = 90° (Given)
- ∴ ΔAOB ≅ ΔAOC (RHS Congruence Rule)
- ∴ ∠B = ∠C and BO = CO (c.p.c.t.)
Worked example
In ΔABC, AB = AC and AO ⊥ BC. If BO = 5 cm, what is the length of CO?
Since ΔAOB ≅ ΔAOC by RHS congruence, corresponding sides BO and CO are equal. Thus, CO = 5 cm.
- 4 cm
- 5 cm — correct
- 6 cm
- 10 cm