In a quadrilateral ABCD, and , then what is equal to?

. In right at B, . Substituting this into the given equation gives . By the converse of the Pythagoras theorem in , the angle opposite to side must be . Thus, .

In a quadrilateral ABCD, \angle B=90^{\circ} and AB^{2}+BC^{2}+CD^{2}-AD^{2}=0, then what is \angle ACD equal to?

A. 30^{\circ}
B. 60^{\circ}
C. 90^{\circ} ✓
D. 120^{\circ}

Correct Answer: C. 90^{\circ}

Explanation

In right \Delta ABC at B, AB^2 + BC^2 = AC^2. Substituting this into the given equation gives AC^2 + CD^2 - AD^2 = 0 \Rightarrow AC^2 + CD^2 = AD^2. By the converse of the Pythagoras theorem in \Delta ACD, the angle opposite to side AD must be 90^\circ. Thus, \angle ACD = 90^\circ.

In a quadrilateral ABCD, \angle B=90^{\circ} and AB^{2}+BC^{2}+CD^{2}-AD^{2}=0, then what is \angle ACD equal to?

Explanation

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