In the figure below, $∠ A C B = 40^{\circ}$ and $B C = A C .$ Also, $E$ is the mid-point of $A B . ∠ A C E$ is equal to

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

No worries! We‘ve got your back. Try BYJU‘S free classes today!

None of these

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A

In $△ A E C$ and $△ B E C,$

$∠ A = ∠ B$ [Angles corresponding to equal sides]

$A E = B E$ [ $E$ is the mid-point of $A B$ ]

$B C = A C$ [Given]

$\Rightarrow △ A E C ≅ △ B E C$ [SAS congruency]

$\Rightarrow ∠ B C E = ∠ A C E$ [CPCTE]

$∠ A C B = ∠ B C E + ∠ A C E = 2 ∠ A C E$

$\Rightarrow ∠ A C E = \frac{∠ A C B}{2} = \frac{40^{\circ}}{2} = 20^{\circ}$

Suggest Corrections

Similar questions

In the figure below, $∠$ C= $40^{\circ}$ and CB=AC. Also, EC is the bisector of AB. $∠$ ACE is equal to _____

In the figure, ABC is an isosceles triangle in which AB = AC. If D and E are the mid-points of sides AB and BC respectively and $∠$ DOE = 120º, find
$∠$ ODE.

In $Δ A B C ≅ Δ A C B, t h e n Δ A B C$ is isosceles with

In triangle ABC, D is a point on AB and E is a point on AC such that DE || BC. If $\frac{A D}{A B}$ = $\frac{A E}{x}$ , Then $x$ is equal to ___.

Q. In the given figure, D is the midpoint of BC, DE ⊥ AB and DF ⊥ AC such that DE = DF. Then, which of the following is true?
(a) AB = AC
(b) AC = BC
(c) AB = BC
(d) None of these

Join BYJU'S Learning Program

In the figure below, ∠ACB=40∘ and BC=AC. Also, E is the mid-point of AB. ∠ACE is equal to

The correct option is A In △AEC and △BEC, ∠A=∠B [Angles corresponding to equal sides] AE=BE [E is the mid-point of AB] BC=AC [Given] ⇒△AEC≅△BEC [SAS congruency] ⇒∠BCE=∠ACE [CPCTE] ∠ACB=∠BCE+∠ACE=2∠ACE ⇒∠ACE=∠ACB2=40∘2=20∘

In the figure below, $∠ A C B = 40^{\circ}$ and $B C = A C .$ Also, $E$ is the mid-point of $A B . ∠ A C E$ is equal to