Triangles ABC and ABD are isosceles with AB=AC=BD, and BD intersects AC at E. If BD⊥AC, then ∠ C+∠ D is

1996 AMC 12 Problem 21

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21.

Triangles

ABC

and

ABD

are isosceles with

AB=AC=BD,

and

\overline{BD}

intersects

\overline{AC}

E.

\overline{BD}\perp\overline{AC},

then

\angle C+\angle D

115^\circ

120^\circ

130^\circ

135^\circ

not uniquely determined

Let

\angle BAC=\alpha

and express

\angle C

using isosceles triangle

ABC

Use

BD\perp AC

to find the vertex angle at

B

of isosceles triangle

ABD

Let

\angle BAC=\alpha.

Since

AB=AC,

\angle C=\frac{180^\circ-\alpha}{2}=90^\circ-\frac{\alpha}{2}.

Because

BD\perp AC,

the angle between

BA

and

BD

90^\circ-\alpha.

In isosceles triangle

ABD,

AB=BD,

so its two base angles are

\begin{aligned} \angle D &=\frac{180^\circ-(90^\circ-\alpha)}2\\ &=45^\circ+\frac{\alpha}{2}. \end{aligned}

Their sum is

135^\circ,

so the correct answer is D.