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.
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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