NCERT Exemplar Solutions
Class 10 - Mathematics - CHAPTER 6: Triangles - Exercise 6.4

Question 12

Step 1: Extend sides \(AB\) and \(DC\) so that they meet at a point \(E\).

Step 2: Since \(\angle A + \angle D = 90^\circ\), the exterior angle at point \(E\) becomes

\(\angle AED = 180^\circ - (\angle A + \angle D) = 180^\circ - 90^\circ = 90^\circ.\)

So, triangle \(AED\) is a right-angled triangle at \(E\).

Step 3: Similarly, angle \(\angle AEC = 90^\circ\) and angle \(\angle BED = 90^\circ\). This means we now have four right-angled triangles: \(\triangle AED, \triangle AEC, \triangle BED, \triangle BEC\).

Step 4 (Applying Pythagoras):

• In \(\triangle AED\): \(AD^2 = AE^2 + DE^2.\)
• In \(\triangle BEC\): \(BC^2 = BE^2 + CE^2.\)
• In \(\triangle AEC\): \(AC^2 = AE^2 + CE^2.\)
• In \(\triangle BED\): \(BD^2 = BE^2 + DE^2.\)

Step 5 (Add equations): Add the last two results:

\(AC^2 + BD^2 = (AE^2 + CE^2) + (BE^2 + DE^2).\)

Rearranging terms: \(AC^2 + BD^2 = (AE^2 + DE^2) + (BE^2 + CE^2).\)

Step 6: From Step 4, we already know:

\(AE^2 + DE^2 = AD^2\) and \(BE^2 + CE^2 = BC^2.\)

Step 7: Substituting these back:

\(AC^2 + BD^2 = AD^2 + BC^2.\)

Final Answer: Hence proved.