Given: △abc, bc> ac, d∈ ac , cd=cb prove: m∠abd is acute

I want you to imagine as you read this or you can draw through the help of my explanation and see yourself:

1↪Draw triangle ABC where BC>AC

2↪D is any point on AC such that CD=CB

3↪Roughly drawing , you can assume CD=CB and and join BD

4↪SO triangle ABC which is a big triangle is divided into Triangles ABD and BDC

5↪See in triangle BDC ,CD=CB so, base angles of isosceles triangle are equal:
<CDB=<CBD = x (assume) which means x is acute angle since CDB and CBD are are in same triangle with same measure and there can't be any two obtuse angle in any traingle. So x must be acute.

6↪Now see in traingle ABD,

<ADB=180-<CDB=180-x=obtuse angle
...check yourself ...just subtract any acute angle from 180 you will get only obtuse angle (ie angle greater than 90)
That means in triangle ABD , one angle ADB is obtuse which means remaining <ABD and < BAD are acute. [PROVED]

❇Main Concept Used Here:
↪In any triangle there can be maximum of one obtuse angle...so remaining two must be acute angle otherwise interior angles sum can't be equal to 180.