euclidean-geometrygeometryplane-geometry
In a parallelogram $ABCD$ the diagonals intersect at $O$. The distances from $O$ to sides $BC$ and $CD$ measure $2$ and $3$ respectively. Calculate the area of the parallelogram if $\angle ABC = 135^o$ (Answer:$24\sqrt2$)
My progess:
$S_{BCD}= \frac{BC.2}{2}=BC\\
S_{OCD} = \frac{CD.3}2=\frac{3CD}{2}\\
S_{BOC}=S_{COD} \implies BC = \frac{3CD}{2}\\
S_{ABCD} = 4BC = 6CD$
…ome detail is missing and I didn't see the given angle function
A cyclic quadrilateral is a quadrilateral whose vertices all lie on a circle. So they are necessarily equidistant from the point $O$, aren't they?
And you have two arcs: $CD=120$, and $BC=60$. So, $BD=180$.
Unbelievable setup! I had great joy of simply angle chasing and discovering and I wish to share with you.
$\triangle BFG$ is the reference triangle. $FP,GQ$ are given its external angle bisectors; let them meet at excenter $E$. Drop perpendiculars from $E$ onto $AB,BC$ and complete the square $ABCE$. Lets observe that $AEC$ is exact copy of $ABC$, so $\angle PEQ = 45^\circ$.
Now the fun begins. $\angle PEG=45^\circ=\angle PCG$, hence $PECG$ is cyclic. $\angle EPG$ being opposite to $\angle ECG$ is a right angle. Similarly $AFQE$ is cyclic, making $\angle FQE$ another right angle. Observe that $FG$ subtends $90^\circ$ at $B,P,Q$. As a result, $B,F,P,Q,G$ lie on same circle $!!$ Its center is $D$, midpoint of $FG$.
Now $DP=DF=$ radius of circle, $\angle DPF = \angle DFP = \angle PFA$ implying that $DP \parallel BA$. Hence $\triangle BPF$ and $\triangle BDF$ have same area. Take away their common area and we get $\triangle FRP$ and $\triangle BRD$ have same area. Similarly $DQ \parallel BC$ resulting in $\triangle GSQ$ and $\triangle BSD$ having same area. Adding the common area of $PRSQ$ to these, we see $FPQG$ and $\triangle BPQ$ have same area.
We do angle chasing one more time to find some lengths. $\angle BPC =$$ \angle PAB + \angle ABP = \angle PBQ + \angle ABP = \angle ABQ $. Therefore $\triangle ABQ \sim \triangle CPB$ by $AA$ similarity. So $$\frac{AB}{PC}=\frac{AQ}{BC} \Rightarrow AB^2=9\times 8 \Rightarrow AB=6\sqrt{2} \Rightarrow AC=12$$
From this it is found $AP=3, PQ=5, QC=4 \, (!)$ Consequently, $$[FPQG]=[BPQ]=\frac{PQ}{AC}\times [ABC]=\frac{5}{12}\times 36=15 \quad \square$$
Best Answer
Hint: drop the altitude $AH$ from $A$ to $DC$. What is the length of $AH$? What are the angles in $\triangle DAH$?Can you calculate the area of parallelogram knowing the base and the height?