Question

In Fig. 14.40, a right triangle BOA is given. C is the mid-point of the hypotenuse AB. Show that it is equidistant from the vertices 0, A and B.

Answer 1

We have to prove that vertices A, B and O are equidistant from C of right triangle BOA where C is the midpoint of AB

• Distance of A from O -: 2a and Distance of B from O -: 2b
• Applying pythagoras theorem, OC² + AC² = OA²                     (1)
• Similarly, OC² + BC² = OB²                                                          (2)
• Applying pythagoras theorem to triangle OAC, OA² + OB² = AB²
• substituting, (2a)² + (2b)² = AB²
• 4a² + 4b² = AB²
• AB = 2√a² +b²
• therefore, AC=BC=√a²+b²                       (3)
• From (1), we get - OC² + a²+b² = 4a²
• OC² = 3a²-b²
• similarly, from (2), we get - OC² + a²+b² = 4b²
• OC² = 3b²-a²
• Hence, 3b²-a² = 3a²-b²
• 4b² = 4a² ;  b² = a²
• Now, from (3) we get, AC=BC= √a²+a² ; a√2
• OC² = 3a²-b² = 3a² - a² = 2a²
• OC = a√2
• Hence, OC=AC=BC
• therefore, C is equidistant from vertices O, A and B