If D is a point on side BC of a triangle ΔABC such that AD = BD = CD, then:

Correct option is A

Given:

D lies on BC

AD = BD = CD

This implies that:

AD is the median to BC

BD = DC (since D is the midpoint)

AD is also perpendicular to BC (as per construction shown)

Solution:

Apply the Pythagoras Theorem in triangles ABD and ACD separately:

In triangle ABD:
AB² = AD² + BD²

In triangle ACD:
AC² = AD² + CD²

Since BD = CD, and AD is common:

Add both equations:
AB² + AC² = AD² + BD² + AD² + CD²
= 2AD² + BD² + CD²

Now since BD = CD = ½ BC
​$$BD^2 + CD^2 = 2 \times \left( \frac{1}{2} BC \right)^2\\ = 2 \times \frac{BC^2}{4} = \frac{BC^2}{2}\\\Rightarrow AB^2 + AC^2 = 2AD^2 + \frac{BC^2}{2}$$​​

From triangle property where AD = ½ BC, and using full simplification as shown in the image:
→ AB² + AC² = BC²

Final Answer: (a) AB² + AC² = BC²