In an Isoceles $△$ $triangle$ $A B C$ $ABC$ ,bisector $C D$ $CD$ of the $∠$ $angle$ $C$ $C$ is equal to the base $B C$ $BC$ .Then the angle between $C D A$ $CDA$ is ?

Question

In an Isoceles $△$ $triangle$ $A B C$ $ABC$ ,bisector $C D$ $CD$ of the $∠$ $angle$ $C$ $C$ is equal to the base $B C$ $BC$ .Then the angle between $C D A$ $CDA$ is ?

1 Answer

Impact of this question

3681 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-02T18:29:36

$m ∠ C D A = 108^{o}$ $m/_CDA=108^o$

Explanation:

enter image source here
Since $A B C$ $ABC$ is isosceles with base $B C$ $BC$ , we know that $A B = A C$ $AB=AC$ . Therefore, $m ∠ A B C = m ∠ A C B$ $m/_ABC=m/_ACB$ . Since we are given that $C D = B C$ $CD=BC$ , we know that $m ∠ C D B = m ∠ C B D$ $m/_CDB=m/_CBD$ . Let $x = m ∠ D C B$ $x=m/_DCB$ . Since $C D$ $CD$ is the bisector of $∠ C$ $/_C$ , we know that $m ∠ B C A = 2 x = m ∠ A B C = m ∠ C D B$ $m/_BCA=2x=m/_ABC=m/_CDB$ . Since $C D B$ $CDB$ is a triangle, we know that $m ∠ C D B + m ∠ C B D + m ∠ B C D = 108^{o} = x + 2 x + 2 x$ $m/_CDB+m/_CBD+m/_BCD=108^o=x+2x+2x$ .
$5 x = 180^{o}$ $5x=180^o$
$:.x=36^o$ .
$:.m/_CDB=2x=72^o$
Since $/_CDB$ and $/_CDA$ are supplementary, $m/_CDB+m/_CDA=180^o$
$:.72^o+m/_CDA=180^o$
$:.m/_CDA=108^o$

Science

Math

Humanities

... and beyond

In an Isoceles $△$ $triangle$ $A B C$ $ABC$ ,bisector $C D$ $CD$ of the $∠$ $angle$ $C$ $C$ is equal to the base $B C$ $BC$ .Then the angle between $C D A$ $CDA$ is ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

In an Isoceles triangle△triangle ABCABCABC,bisector CDCDCD of the angle∠angle CCC is equal to the base BCBCBC.Then the angle between CDACDACDA is ?

1 Answer

Explanation:

Related questions

Impact of this question

In an Isoceles $△$ $triangle$ $A B C$ $ABC$ ,bisector $C D$ $CD$ of the $∠$ $angle$ $C$ $C$ is equal to the base $B C$ $BC$ .Then the angle between $C D A$ $CDA$ is ?