You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Mathematics

Similarity of Right-angled Triangles

Question 11 I...

Question

Question 11
In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD $⊥$ BC and EF $⊥$ AC, prove that $Δ A B D \sim Δ E C F$ .

Open in App

Solution

It is given that ABC is an isosceles triangle.
$∴ A B = A C$
$\Rightarrow ∠ A B D = ∠ E C F$ ........(i)
In $Δ A B D a n d Δ E C F$ ,
$∠ A D B = ∠ E F C (E a c h 90^{\circ})$
$∠ A B D = ∠ E C F$ (from (i))
$∴ Δ A B D \sim Δ E C F$ (By using AAA similarity criterion)

Suggest Corrections

Similar questions

Question 11
In the following figure, E is a point on side CB produced, of an isosceles triangle ABC, with AB = AC. If AD $⊥$ BC and EF $⊥$ AC, prove that $Δ A B D \sim Δ E C F$ .

Q. In the following figure, E is a point on side CB produced, of an isosceles triangle ABC, with AB = AC. If AD

⊥

BC and EF

⊥

AC, prove that

Δ A B D \sim Δ E C F

Q. E is a point on side CB produced of an isosceles triangle ABC with AB=AC. If

A D ⊥ B C

and

E F ⊥ A C

, prove that

Δ A B D \sim Δ E C F .

In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ∼ ΔECF

Q. In given figure

E

is a point on side

C B

produced of an isosceles triangle

A B C

with

A B = A C .

A D ⊥ B C

and

E F ⊥ A C,

prove that

△ A B D \sim △ E C F

Join BYJU'S Learning Program

Question 11 In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD∼ΔECF.

It is given that ABC is an isosceles triangle. ∴AB=AC ⇒∠ABD=∠ECF........(i) In ΔABD and ΔECF, ∠ADB=∠EFC(Each 90∘) ∠ABD=∠ECF (from (i)) ∴ΔABD∼ΔECF (By using AAA similarity criterion)

Question 11
In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD $⊥$ BC and EF $⊥$ AC, prove that $Δ A B D \sim Δ E C F$ .

It is given that ABC is an isosceles triangle.
$∴ A B = A C$
$\Rightarrow ∠ A B D = ∠ E C F$ ........(i)
In $Δ A B D a n d Δ E C F$ ,
$∠ A D B = ∠ E F C (E a c h 90^{\circ})$
$∠ A B D = ∠ E C F$ (from (i))
$∴ Δ A B D \sim Δ E C F$ (By using AAA similarity criterion)