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Point X is 3m from A, 4m from B and 5m from C. Point X is inside the triangle formed by ABC. If AB = BC and angle B is right angle. Find the length of side AB.
I have come up with the following figure and aware that this is an isosceles right triangle, unfortunately i'm not sure what to do next.
Assign coordinates to the 4 given points. B = $(0,0)$, C = $(L,0)$, A = $(0,L)$, X = $(x, y)$.
Apply Pythagoras to the three line segments.
$$x^2 +y^2 = 16$$
$$(L-x)^2 + y^2 = 25$$
$$x^2 + (L-y)^2 = 9$$
Combining 1. and 2. gives $L^2 - 2Lx = 9$ and therefore $x = (L^2-9)/(2L)$
Combining 1. and 3. gives $L^2 - 2Ly = -7$ and therefore $y = (L^2 +7)/(2L)$
You can now substitute these expressions for $x$ and $y$ into equation 1 and solve for $L$. The result is $L = \sqrt{17 + \sqrt{224}}$, which is approximately 5.6539.
Reflect point $X$ across the sides of $\triangle ABC$. Observe, $$\begin{align*} \angle DAF&=\angle DAX+\angle FAX=2(\angle BAX+\angle CAX)=90^{\circ}\;\implies DF=3\sqrt{2}\\ \angle DCE&=\angle DCX+\angle ECX=2(\angle ACX+\angle BCX)=90^{\circ}\;\implies DE=5\sqrt{2}\\ \angle FBE&=\angle FBX+\angle EBX=2(\angle ABX+\angle CBX)=180^{\circ}\implies F,\;B,\;E \;\;\text{are collinear.} \end{align*} $$ Using the cosine rule, in $\triangle EDF,$ $$\begin{align*} EF^2&=DE^2+DF^2-2\cdot DE\cdot DF\cdot\cos\angle EFD \\ 64&=18+50-60\cdot\cos\angle EFD \\ \cos \angle EFD&=\frac{1}{15} \implies \sin\angle EFD=\frac{4\sqrt{14}}{15} \end{align*} $$ Using the cosine rule, in $\triangle ADC,$ $$\begin{align*} AC^2 &=AD^2+CD^2-2\cdot AD\cdot CD\cdot \cos\angle ADC\\ &=9+25-30\cos(90^{\circ}+\angle EFD)\\ &=34+30\cdot\sin\angle EFD\\ &=34+8\sqrt{14} \\ \\ \therefore\; AB&=\sqrt{\frac{AC^2}{2}}=\boxed{\sqrt{17+4\sqrt{14}}\approx 5.6539} \end{align*}$$
Comment
To construct the triangle in Geogebra I drew three concentric circles radii 3,4,5 for A,B,C respectively with X as center. Constructed a slidable right angle with 90 deg vertex at B. Drawn arc with radius 5.654 units from B to locate A and C as shown.
However got a different location of X inside the triangle with hypotenue AC almost 8 units long and X is almost on side AC. What may be my construction error?
