Respuesta :

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.

The point D is located along the line CA and therefore forms the triangles

ΔBCD and ΔABD.

m∠BAC = m∠CBA + 2 × m∠ABD

m∠BAC < 180°

∴ m∠ABD < 90°

Therefore;

m∠ABD is acute

Detailed reasons:

The given parameters are;

In triangle ΔABC

Length of segment BC  > Length of segment AC

Point D is a point on the line AC

Length of segment CD = Length of segment CB

Required:

To prove that m∠ABD is acute.

Solution:

m∠CBD = m∠CDB by base angles of an isosceles triangle

m∠BAC = m∠CDB + m∠ABD by exterior angle of a triangle theorem

m∠BAC = m∠CBD + m∠ABD by substitution property

m∠CBD = m∠CBA + m∠ABD by angle addition property

m∠BAC = m∠CBA + m∠ABD + m∠ABD = m∠CBA + 2 × m∠ABD

m∠BAC = m∠CBA + 2 × m∠ABD

m∠BAC < 180°

m∠CBA > 0

[tex]m\angle ABD = \dfrac{m\angle BAC - m\angle CBA }{2}[/tex]

Therefore;

[tex]m\angle ABD < \dfrac{180^{\circ} - 0}{2} = 90^{\circ}[/tex]

m∠ABD < 90°

Therefore; m∠ABD is an acute angle, by definition of acute angles.

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