In the figure given $Δ A B C$ is a right isosceles triangle with right angle at C CD is a parallel to AB and BD = BA. The degree measure of $∠ D B C$ equals:

10^{o}

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15^{o}

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20^{o}

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25^{o}

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Solution

The correct option is B $15^{o}$
Draw the $⊥$ lines $B E$ and $C F$ .
Since, $A C = C B ⟹ C F$ bisects $A B$
$\Rightarrow C F = F B = \frac{A B}{2} \Rightarrow B E C F$ is a square.
$B D = A B = B E = C F = \frac{B D}{2}$
$∠ D = 30^{o}, ∠ D B E = 60^{o}$
$∠ D B C = 60 + 45 = 105^{o}$
In case of $C D$ in direction of $B A$
$\Rightarrow ∠ D B C = 45^{o} - 30^{o} = 15^{o}$

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In the figure given ΔABC is a right isosceles triangle with right angle at C CD is a parallel to AB and BD = BA. The degree measure of ∠DBC equals:

The correct option is B 15oDraw the ⊥ lines BE and CF.Since, AC=CB⟹CF bisects AB⇒CF=FB=AB2⇒BECF is a square.BD=AB=BE=CF=BD2∠D=30o,∠DBE=60o∠DBC=60+45=105oIn case of CD in direction of BA⇒∠DBC=45o−30o=15o

In the figure given $Δ A B C$ is a right isosceles triangle with right angle at C CD is a parallel to AB and BD = BA. The degree measure of $∠ D B C$ equals: