In ΔABC, AD is the perpendicular bisector of BC. Show that ΔABC is an isosceles triangle in which AB=AC.
Solution:
In ΔABD and ΔACD, we have DB=DC Given ∠ADB=∠ADC since AD⊥BC AD=AD Common ∴ by SAS criterion of congruence, we have. ΔABD ≅ ΔACD ⇒ AB=AC Since corresponding parts of congruent triangles are equal Hence, ΔABC is isosceles.