Triangle ABC is equilateral. CAD and ABE are straight lines and AD = BE. Show that triangle ABD and triangle BCE are congruent. State your reasons clearly. The diagram shows a straight line passing through the points B (3, 4) and C (7, 6). Line BC cuts the x-axis at A. Find the area of triangle OAB. Answer (b)(i)... units^2 Another point D is such that ΔABO is similar to ΔACD. Find the coordinates of point D. Answer (b)(ii) (... , ... ) Find the numerical value of area of OBCD / area of ΔACD. Answer (b)(iii)...

1. State the equal side pair from the equilateral triangle

Since $\triangle ABC$ is equilateral,
$AB=BC$
and
$\angle BAC=\angle ABC=60^\circ.$

2. Find the included angles in the two triangles

Since $CAD$ is a straight line,
$\angle BAD=180^\circ-\angle BAC=180^\circ-60^\circ=120^\circ.$
Since $ABE$ is a straight line,
$\angle CBE=180^\circ-\angle ABC=180^\circ-60^\circ=120^\circ.$
Hence,
$\angle BAD=\angle CBE.$

3. Apply the SAS congruence test

Also, it is given that
$AD=BE.$
Therefore, in $\triangle ABD$ and $\triangle BCE$:
$AB=BC,$
$AD=BE,$
$\angle BAD=\angle CBE.$
The equal angle is included between the two equal sides in each triangle. Hence,
$\triangle ABD \cong \triangle BCE \quad \text{(SAS)}.$

1. Find the gradient of line $BC$

For $B(3,4)$ and $C(7,6)$,
$m=\frac{6-4}{7-3}=\frac{2}{4}=\frac{1}{2}.$

2. Find the x-intercept $A$ of line $BC$

Using point $B(3,4)$,
$y-4=\frac{1}{2}(x-3).$
At the x-axis, $y=0$.
$0-4=\frac{1}{2}(x-3)$
$-8=x-3$
$x=-5.$
So $A=(-5,0)$.

3. Calculate the area of triangle $OAB$

Since $O=(0,0)$ and $A=(-5,0)$, $OA=5$.

The perpendicular height from $B(3,4)$ to the x-axis is $4$.

$\text{Area of }\triangle OAB=\frac{1}{2}\times 5\times 4=10.$

1. Use the similarity scale factor to find the position of $D$

From part (i), $A=(-5,0)$. Also $B=(3,4)$, $C=(7,6)$ and $O=(0,0)$.

Since $\triangle ABO \sim \triangle ACD$,
$A \leftrightarrow A,\quad B \leftrightarrow C,\quad O \leftrightarrow D.$
Now
$\overrightarrow{AB}=(3-(-5),4-0)=(8,4),$
$\overrightarrow{AC}=(7-(-5),6-0)=(12,6).$
So the scale factor from $\triangle ABO$ to $\triangle ACD$ is
$\frac{AC}{AB}=\frac{12}{8}=\frac{3}{2}.$
Therefore
$\overrightarrow{AD}=\frac{3}{2}\overrightarrow{AO}=\frac{3}{2}(5,0)=(7.5,0).$
Hence
$D=(-5,0)+(7.5,0)=(2.5,0).$

1. Find the area of quadrilateral $OBCD$ using coordinates

Using $O(0,0)$, $B(3,4)$, $C(7,6)$ and $D(2.5,0)$,

$\text{Area of }OBCD=\frac12\left|0(4)+3(6)+7(0)+2.5(0)-[0(3)+4(7)+6(2.5)+0(0)]\right|$

$=\frac12|18-43|=12.5$

2. Find the area of $\triangle ACD$ and calculate the required ratio

From earlier parts, $A=(-5,0)$ and $D=(2.5,0)$.

So

$AD=2.5-(-5)=7.5$

The perpendicular height from $C(7,6)$ to the $x$-axis is $6$.

$\text{Area of }\triangle ACD=\frac12(7.5)(6)=22.5$

Hence

$\frac{\text{area of }OBCD}{\text{area of }\triangle ACD}=\frac{12.5}{22.5}=\frac{5}{9}$