Question

D is a point on the side BC of \(\Delta ABC\) such that \(\angle CAB = \angle CDA\). Show that \(CA^2 = CB \times CD\).

Hint:

To avoid mistakes when writing down the ratios of corresponding sides of similar triangles, always trace the letters directly from your similarity statement:
If \(\Delta \textbf{C}\textbf{A}B \sim \Delta \textbf{C}\textbf{D}A\):
- The first two letters of both triangles give the ratio: \(\frac{CA}{CD}\)
- The first and third letters of both triangles give the ratio: \(\frac{CB}{CA}\)
Setting these equal directly yields the correct ratio without having to look back at the geometric diagram!

Answer 1

Step 1: Identify two triangles sharing an angle.
Look at triangle CAB and triangle CDA, both share the angle at C, since D lies on BC, the angle ACB and angle DCA are literally the same angle.
Step 2: Use the given angle equality.
We are told \(\angle CAB = \angle CDA\), so between the two triangles we now have two pairs of equal angles: \(\angle CAB = \angle CDA\) and \(\angle ACB = \angle DCA\) (the common angle).
Step 3: Apply the AA similarity criterion.
Since two angles of triangle CAB equal two angles of triangle CDA, by the AA criterion: \[ \Delta CAB \sim \Delta CDA \] with C matching C, A matching D, and B matching A.
Step 4: Translate the similarity into a side ratio.
Matching sides in this order gives \(\frac{CA}{CD} = \frac{CB}{CA}\), and cross-multiplying: \[ CA \times CA = CB \times CD \implies CA^2 = CB \times CD \] Hence, it is proved that \(CA^2 = CB \times CD\).