# 2002 AMC 12A Problems/Problem 23

## Problem

In triangle $ABC$, side $AC$ and the perpendicular bisector of $BC$ meet in point $D$, and $BD$ bisects $\angle ABC$. If $AD=9$ and $DC=7$, what is the area of triangle $ABD$? $\text{(A)}\ 14 \qquad \text{(B)}\ 21 \qquad \text{(C)}\ 28 \qquad \text{(D)}\ 14\sqrt5 \qquad \text{(E)}\ 28\sqrt5$

## Solution 1 $[asy] unitsize(0.25 cm); pair A, B, C, D, M; A = (0,0); B = (88/9, 28*sqrt(5)/9); C = (16,0); D = 9/16*C; M = (B + C)/2; draw(A--B--C--cycle); draw(B--D--M); label("A", A, SW); label("B", B, N); label("C", C, SE); label("D", D, S); [/asy]$ Looking at the triangle $BCD$, we see that its perpendicular bisector reaches the vertex, therefore implying it is isosceles. Let $x = \angle C$, so that $B=2x$ from given and the previous deducted. Then $\angle ABD=x, \angle ADB=2x$ because any exterior angle of a triangle has a measure that is the sum of the two interior angles that are not adjacent to the exterior angle. That means $\triangle ABD$ and $\triangle ACB$ are similar, so $\frac {16}{AB}=\frac {AB}{9} \Longrightarrow AB=12$.

Then by using Heron's Formula on $ABD$ (with sides $12,7,9$), we have $[\triangle ABD]= \sqrt{14(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$.

## Solution 2

Let M be the point of the perpendicular bisector on BC. By the perpendicular bisector theorem, $BD = DC = 7$ and $BM = MC$. Also, by the angle bisector theorem, $\frac {AB}{BC} = \frac{9}{7}$. Thus, let $AB = 9x$ and $BC = 7x$. In addition, $BM = 3.5x$.

Thus, $\cos\angle CBD = \frac {3.5x}{7} = \frac {x}{2}$. Additionally, using the Law of Cosines and the fact that $\angle CBD = \angle ABD$, $81 = 49 + 81x^2 - 2(9x)(7)\cos\angle CBD$

Substituting and simplifying, we get $x = 4/3$

Thus, $AB = 12$. We now know all sides of $\triangle ABD$. Using Heron's Formula on $\triangle ABD$, $\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$

"Note:-you could also drop a perpendicular from D to AB at point let say,F then BF = 3.5x by pyathgoras theorem we can find DF and (AB ×DF )÷2 is our answer"

## Solution 3

Note that because the perpendicular bisector and angle bisector meet at side $AC$ and $CD = BD$ as triangle $BDC$ is isosceles, so $BD = 7$. By the angle bisector theorem, we can express $AB$ and $BC$ as $9x$ and $7x$ respectively. We try to find $x$ through Stewart's Theorem. So $16(7^2+9\cdot7) = (7x)^2 \cdot 9 + (9x)^2 \cdot 7$ $16(49+63) = (49 \cdot 9 + 81 \cdot 7)x^2$ $16(49+63) = 9(49+63) \cdot x^2$ $x^2 = \frac{16}{9}$ $x=\frac{4}{3}$

We plug this to find that the sides of $\triangle ABD$ are $12,7,9$. By Heron's formula, the area is $\sqrt{(14)(2)(7)(5)} = 14\sqrt5 \Longrightarrow \boxed{\text{D}}$. ~skyscraper

## Solution 4 $[asy] size(12cm, 12cm); pair A, B, C, D, M, N; A = (0,0); B = (88/9, 28*sqrt(5)/9); C = (16,0); D = 9/16*C; M = (B + C)/2; N = (6,4.27); draw(A--B--C--cycle); draw(B--D--M); draw(D--N--B); label("A", A, SW); label("B", B, N); label("C", C, SE); label("D", D, S); label("M", M, NE); label("N", N, NW); draw(rightanglemark(B, M, D), linewidth(.5)); draw(rightanglemark(A, N, D), linewidth(.5)); [/asy]$

Draw $DN$ such that $DN \bot AB$, $\triangle BND \cong \triangle BMD$ $\angle ACB = \angle DBC = \angle ABD$, $\triangle ABD \sim \triangle ACB$ by $AA$ $\frac{AB}{AD} = \frac{AC}{AB}$, $AB^2 = AD \cdot AC = 9 \cdot 16$, $AB = 12$

By the Angle Bisector Theorem, $\frac{BC}{AB} = \frac{CD}{AD}$ $\frac{BC}{12} = \frac{7}{9}$ $BC = \frac{28}{3}$, $CM = \frac{14}{3}$, $DN = DM = \sqrt{CD^2 - CM^2} = \frac{7 \sqrt{5} }{3}$ $[ABD] = \frac12 \cdot AB \cdot DN = \frac12 \cdot 12 \cdot \frac{7 \sqrt{5} }{3} = \boxed{\textbf{(D) } 14 \sqrt{5} }$

## Solution 5 (Trigonometry)

Let $\angle ACB = \theta$, $\angle DBC = \theta$, $\angle ABD = \theta$, $\angle ADB = 2 \theta$, $\angle BAC = 180^\circ - 3 \theta$ $[ABD] = \frac12 \cdot AD \cdot BD \cdot \sin \angle ADB = \frac12 \cdot 9 \cdot 7 \cdot \sin 2\theta$

By the Law of Sines we have $\frac{ \sin \angle BAC }{BD} = \frac{ \sin \angle ABD }{AD}$ $\frac{ \sin(180^\circ - 3 \theta) }{7} = \frac{ \sin \theta }{ 9 }$ $\frac{ \sin 3 \theta }{7} = \frac{ \sin \theta }{ 9 }$

By the Triple-angle Identities, $\sin 3 \theta = 3 \sin \theta - 4 \sin^3 \theta$ $3 - 4 \sin^2 \theta = \frac79$, $36 \sin^2 \theta = 20$, $\sin^2 \theta = \frac59$ $\sin \theta = \frac{\sqrt{5}}{3}$, $\cos \theta = \frac{2}{3}$

By the Double Angle Identity $\sin 2 \theta = 2 \sin \theta \cos \theta = 2 \cdot \frac{\sqrt{5}}{3} \cdot \frac{2}{3} = \frac{ 4 \sqrt{5} }{9}$ $[ABD] = \frac12 \cdot 9 \cdot 7 \cdot \frac{ 4 \sqrt{5} }{9} = \boxed{\textbf{(D) } 14 \sqrt{5} }$

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 