D is a point on side BC of AABC such that AD = AC. Show that AB > AD.
In an isosceles triangle ABC with AB - ACD and are points on Bench
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In quadrilateral ABCD we have
AC = AD
and AB being the bisector of ∠A.
Now, in ΔABC and ΔABD,
AC = AD
[Given]
AB = AB
[Common]
∠CAB = ∠DAB [∴ AB bisects ∠CAD]
∴ Using SAS criteria, we have
ΔABC ≌ ΔABD.
∴ Corresponding parts of congruent triangles (c.p.c.t) are equal.
∴ BC = BD.
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