In an obtuse-angled $\Delta ABC$ (obtuse at $B$), $AD$ is perpendicular to $CB$ produced. Prove that $AC^2 = AB^2 + BC^2 + 2BC \times BD$.

Step 1: Identify Right-Angled Triangles

Given that $AD$ is perpendicular to $CB$ produced, it means that $AD$ forms a right angle with the line segment $CB$ (extended). This creates two right-angled triangles: $\triangle ADB$ and $\triangle ADC$.

Step 2: Apply Pythagoras Theorem in $\triangle ADB$

In the right-angled triangle $\triangle ADB$, $AB$ is the hypotenuse. According to the Pythagoras theorem, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

Step 3: Apply Pythagoras Theorem in $\triangle ADC$

Similarly, in the right-angled triangle $\triangle ADC$, $AC$ is the hypotenuse. Applying the Pythagoras theorem, we get $AC^2 = AD^2 + CD^2$. Let's call this Equation 2.

Step 4: Express CD in terms of BC and BD

From the figure, we can observe that the segment $CD$ is the sum of the segments $CB$ and $BD$. Since $CB$ is produced to $D$, $C$, $B$, and $D$ are collinear, and $B$ lies between $C$ and $D$.

Step 5: Substitute CD into Equation 2

Substitute the expression for $CD$ from the previous step into Equation 2. This expands the term $CD^2$ into $(CB + BD)^2$.

Step 6: Expand and Substitute

Expand $(CB + BD)^2$ using the identity $(a+b)^2 = a^2 + b^2 + 2ab$. Then, rearrange the terms and substitute $AD^2 + BD^2$ with $AB^2$ from Equation 1.

Step 7: Final Proof

By substituting $AD^2 + BD^2 = AB^2$ (from Equation 1) into the expanded Equation 2, we arrive at the desired result.

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