$ABCD$ is a convex quadrilateral with its side lengths $AB = 5, BC = 17, CD = 7, AD = 25,$ and $∠ABC + ∠BCD = 270°.$ Find the area of the quadrilateral $ABCD$.
$ABCD$ is a convex quadrilateral with its side lengths $AB = 5, BC = 17, CD = 7, AD = 25,$ and $∠ABC + ∠BCD = 270°.$
geometry
Related Solutions
Let angle bisector for $\angle BCD$ meet $AB$ at $F$ (so we have to prove that $DF$ is angle bisector for $\angle ADC$), then $$\angle BCF = FCD = \alpha\;\;\;\;{\rm and }\;\;\;\;\;\angle BAD = 180^{\circ} -2\alpha$$ and let $E$ on $AB$ be such that $BE = BC$, then $AE = AD$ and $$\angle FED = \angle AED = \angle ADE = \alpha$$ so $\angle FED = \angle FCD = \alpha $ and thus $CDFE$ is cyclic!
Now, if $\angle FDC = \beta$ then $\angle BEC = \angle ECB = \beta$, so $$\angle 180^{\circ} -2\beta \implies \angle ADC = 2\beta$$ and thus $DF$ is also angle bisector but for $\angle ADC$ and we are done.
ADEF is cyclic, denote by $a$ its interior angle in $A$. Let $x$ be the common value of the two angles $\angle(BAE)$ and $\angle(FDC)$.
Then the angle in $A$ of $ABCD$ is $a+x$, and the exterior angle in $C$ of it is the exterior angle of the $\Delta CDF$ with angles $a$ (in $F$) and $x$ (in $D$), which is thus also $a+x$.
So $ABCD$ is cyclic.
Best Answer
Finally!!
In $\triangle PAD, \; \angle{P} = 90^{\circ}$. From this I got $$x^2 + y^2 = 17^2 \; \text{and} \; 5x + 7y = 131$$ From these two equations I got $x = 15$ and $y = 8$, only possible solution.
Area of $ABCD = 90 \; \text{units}^2$