Question

please solve this






ohk rohan
btw nice to meet you ​

Answer 1

In the diagram , ABC is a triangle and the line segment DE is parallel to BC . Let us check triangle ABC and triangle ADE for similarity .

We know that corresponding angles on two parallel lines are equal .

Hence , angle ADE is equal to angle ABC and angle AED is equal to angle ACB .

$\angle$ADE = $\angle$ABC .. ( 1 )

$\angle$AED = $\angle$ACB .. ( 2 )

$\angle$A is common between both the triangles.

Hence , we have :

$\angle$EAD = $\angle$BAC .. ( 3 )

From equation ( 1 ) , equation ( 2 ) and equation ( 3 ) , the triangles ABC and ADE are similar by Angle Angle Angle ( AAA ) criterion .

Hence , we have : $\triangle$ABC $\sim$ $\angle$ADE.

We know that the properties of similar triangles states that the ratio of corresponding sides of similar triangles are equal .

For triangle ABC and triangle ADE , we have AB and AD , AC and AE , BC and DE as corresponding sides .

Hence , the ratio of AB by AD is equal to the ratio of AC by AE is equal to the ratio of BC by DE .

Let the side CE be x.

AC = 9+x

$\dfrac{AD}{AB} = \dfrac{AE}{AC} \\ \\ \dfrac{6}{14} = \dfrac{9}{9 + x} \\ \\ 54 + 6x = 126 \\ \\ 6x = 72 \\ \\ x = 12 \: cm$

Therefore, CE = 12 cm.

Hope it helps u

Have a great day ahead

Take care and keep smiling

Annyeong