In given figure if AC = BC, $∠$ DCA = $∠$ ECB and $∠$ DBC = $∠$ EAC, then DC = __.

Open in App

Solution

We have,
$∠$ DCA = $∠$ ECB
$\Rightarrow$ $∠$ DCA + $∠$ ECD = $∠$ ECB + $∠$ ECD
$\Rightarrow$ $∠$ ECA = $∠$ DCB .......(i)

Now, in $Δ$ s DBC and EAC, we have
$∠$ DCB = $∠$ ECA [From (i)]
BC = AC [Given]
and $∠$ DBC = $∠$ EAC [Given]

So, by ASA criterion of congruence
$Δ$ DBC $≅$ $Δ$ EAS
$\Rightarrow$ DC = EC [ Correcponding parts of congruent triangles are equal ]

Suggest Corrections

120

Similar questions

∠ D C A = ∠ E C B

∠ D B C = ∠ E A C

Δ D B C ≅ Δ E A C

D C = E C

A B = A C

∠ D B C = ∠ E C B = 90^{\circ}

A D = A E

A B = A C

∠ D B C = ∠ E C B = 90^{\circ}

B D = C E

State true or false:
In the give figure, $A C = A E, A B = A D$ and $∠ B A D = ∠ E A C$ .

B C = D E

A. True

B. False

Join BYJU'S Learning Program